Book 1. Foundations of Olympiad Algebra

Let \(U=x^2+2x+3\), \(V=x^2-3\). The expression equals \((U-V)(U+V)\).

We have \(U-V=2x+6=2(x+3)\), \(U+V=2x^2+2x=2x(x+1)\). Hence the factorization is \(4x(x+1)(x+3)\).

Use \(u^3+v^3=(u+v)(u^2-uv+v^2)\), with \(u=2a\), \(v=3b\).

Thus \(8a^3+27b^3=(2a+3b)(4a^2-6ab+9b^2)\).

Group: \[ a^2b-a^2c+b^2c-bc^2=a^2(b-c)+bc(b-c). \] Factor out \(b-c\):

\((b-c)(a^2+bc)\).

We have \(x^2+y^2=7^2-2\cdot10=29\).

Also \(x^3+y^3=(x+y)(x^2-xy+y^2)=7(29-10)=133\).

If \(n\) is odd, then \(n^3\) is odd, and \[ n^3=\left(\frac{n^3+1}{2}\right)^2-\left(\frac{n^3-1}{2}\right)^2. \]

If \(n\) is even, write \(n=2m\). Then \(n^3=8m^3\). Take \(A=2m^3+1\), \(B=2m^3-1\). Then \(A^2-B^2=(A-B)(A+B)=2\cdot4m^3=8m^3=n^3\).

Transform: \[ x^4+4y^4=x^4+4x^2y^2+4y^4-4x^2y^2=(x^2+2y^2)^2-(2xy)^2. \] Hence \[ x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2). \]

For \(x>1\), \(y\ge1\), both factors are greater than \(1\). Thus the number is composite.

By the sum of cubes identity, \[ x^3+y^3=(x+y)^3-3xy(x+y). \] Since \(x+y=1\), we get \(x^3+y^3=1-3xy\).

Thus \(x^3+y^3+3xy=1\).

From \((a+b+c)^2=0\), we get \(a^2+b^2+c^2+2ab+2bc+2ca=0\). Hence \(a^2+b^2+c^2=-2(ab+bc+ca)\).

Also \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\). Thus \(a^3+b^3+c^3=3abc\).

We have \(n^2-1=(n-1)(n+1)\). For \(n>2\), both factors are greater than \(1\): \(n-1>1\), \(n+1>1\).

Therefore \(n^2-1\) is a product of two integers greater than \(1\), so it is composite.

Squaring, \[ \left(x+\frac1x\right)^2=x^2+2+\frac1{x^2}=9. \] Hence \(x^2+\frac1{x^2}=7\).

Write \[ a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}). \] Expanding shows that all intermediate terms cancel.

Hence \(a^n-b^n\) has factor \(a-b\).

Let \(s=x+y=4\), \(p=xy\). Then \(x^3+y^3=s^3-3ps=64-12p\).

Since \(64-12p=28\), we get \(p=3\). Thus \(x,y\) are roots of \(t^2-4t+3=0\), so \(t=1\) or \(t=3\). The pairs are \((1,3)\), \((3,1)\).

By the identity, \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \] The left side is zero, and \(a+b+c>0\), so the second factor is zero.

But \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \] A sum of squares is zero only when \(a=b=c\).

We have \(x^4+2x^2y^2+y^4=(x^2+y^2)^2\). Thus the expression is \((x^2+y^2)^2-4^2\).

By difference of squares, it factors as \((x^2+y^2-4)(x^2+y^2+4)\). Over the reals, the second factor does not split further.

From \(a+b+c=0\), \(a^2+b^2+c^2=-2(ab+bc+ca)\). Square: \[ (a^2+b^2+c^2)^2=4(ab+bc+ca)^2. \]

The left side is \(a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)\). The right side is \(4(a^2b^2+b^2c^2+c^2a^2+abc(a+b+c))\), and the last term is zero. This gives the required equality.

Use \(n^4+4=(n^2-2n+2)(n^2+2n+2)\).

If \(n>1\), both factors are greater than \(1\), so the number is composite. For \(n=1\), \(1^4+4=5\), which is prime. Answer: \(n=1\).

It is enough to prove \(a^2+b^2+c^2-ab-bc-ca\ge0\). Multiply by \(2\): \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \]

The right side is non-negative, so the inequality is proved.

Let the expression be \(E\). It has degree \(3\), and if \(a=b\), then \(E=0\), so \(a-b\) is a factor. Similarly, \(b-c\) and \(c-a\) are factors.

Since \((a-b)(b-c)(c-a)\) also has degree \(3\), only the sign remains. Substitute \(a=1\), \(b=2\), \(c=3\): \(E=1(-1)+4(2)+9(-1)=-2\). But \((a-b)(b-c)(c-a)=(-1)(-1)(2)=2\). Hence \(E=-(a-b)(b-c)(c-a)\).

Compute: \[ (x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3. \] By the conditions this is \(3-6+3=0\).

A sum of squares is zero only when each square is zero. Hence \(x=y=z=1\).

We have \((x-y)(x+y)=15\). The two factors have the same parity because their sum is \(2x\), even. Since \(15\) is odd, both are odd.

The possible pairs are \((1,15)\), \((3,5)\), \((-1,-15)\), \((-3,-5)\). Solving \(x-y=u\), \(x+y=v\), we get \((x,y)=(8,7),(4,1),(-8,-7),(-4,-1)\).

Since \(a+b+c=0\), the identity gives \(a^3+b^3+c^3=3abc\).

Given \(abc=2\), we get \(a^3+b^3+c^3=6\).

Consider the difference: \[ a^4+b^4-a^3b-ab^3=a^3(a-b)-b^3(a-b)=(a-b)(a^3-b^3). \] Also \(a^3-b^3=(a-b)(a^2+ab+b^2)\).

So the difference is \((a-b)^2(a^2+ab+b^2)\). The first factor is non-negative, and the second is \(\left(a+\frac b2\right)^2+\frac34b^2\ge0\). Hence the inequality is proved.

We have \(n^4+4n^2+4=(n^2+2)^2\). A square of an integer greater than \(1\) is never prime.

Here \(n^2+2\ge3\), so \((n^2+2)^2\) has a divisor \(n^2+2\), different from \(1\) and from the number itself. Hence there are no solutions.

Let \(s=x+y\), \(p=xy\). From \(x^2+y^2=s^2-2p=10\), we get \(p=\frac{s^2-10}{2}\).

Also \(x^3+y^3=s^3-3ps=10\). Substitute \(p\): \[ s^3-\frac{3s(s^2-10)}{2}=10. \] Multiplying by \(2\), \(2s^3-3s^3+30s=20\), so \(s^3-30s+20=0\) after changing sign.

Additionally, real \(x,y\) require \(s^2-4p=s^2-2(s^2-10)=20-s^2\ge0\). Thus the possible values of \(x+y\) are precisely the real roots of \(s^3-30s+20=0\) lying in \([-\sqrt{20},\sqrt{20}]\).

We have \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \] Since \(a,b,c\ge0\), the first factor \(a+b+c\ge0\).

The second factor is non-negative because \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \] Hence the whole difference is non-negative. Equality occurs when \(a+b+c=0\), which gives \(a=b=c=0\), or when the second factor is zero, which gives \(a=b=c\). Thus equality holds exactly when \(a=b=c\).

Factor out \(3xy\): \(6x^3y-9x^2y^2+3xy^3=3xy(2x^2-3xy+y^2)\).

Then \(2x^2-3xy+y^2=(2x-y)(x-y)\). Hence the full factorisation is \(3xy(2x-y)(x-y)\).

We have \(ab+ac+bd+cd=a(b+c)+d(b+c)\).

Factoring out \(b+c\), we get \((a+d)(b+c)\).

The two numbers are \(p\) and \(q\). Therefore \(x^2-(p+q)x+pq=(x-p)(x-q)\).

Let \(U=2a+b\), \(V=a-2b\). Then \(U-V=a+3b\), \(U+V=3a-b\).

Thus the expression equals \((a+3b)(3a-b)\).

We have \(a^2+2ab+b^2-c^2=(a+b)^2-c^2\).

This is a difference of squares: \((a+b-c)(a+b+c)\).

\(x^4+4y^4=x^4+4x^2y^2+4y^4-4x^2y^2=(x^2+2y^2)^2-(2xy)^2\).

Hence \(x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2)\).

For odd \(m\), \(a^m+b^m=(a+b)(a^{m-1}-a^{m-2}b+\cdots-ab^{m-2}+b^{m-1})\).

With \(m=2n+1\), the factor \(a+b\) appears, so the divisibility is proved.

Let \(P(x)=x^3+kx^2-5x-2k\). The condition is equivalent to \(P(2)=0\).

\(P(2)=8+4k-10-2k=2k-2\), hence \(k=1\).

Then \(P(x)=x^3+x^2-5x-2=(x-2)(x^2+3x+1)\).

\(x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2\).

Thus the factorisation is \((x^2-x+1)(x^2+x+1)\).

We have \(n^4+4n^2+3=(n^2+1)(n^2+3)\).

For positive \(n\), both factors are greater than \(1\), so the number is composite.

By the identity, \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).

The second factor is an integer when \(a,b,c\) are integers. Therefore the expression is divisible by \(a+b+c\).

\(x^2y^2-9z^4=(xy)^2-(3z^2)^2\).

Hence \(x^2y^2-9z^4=(xy-3z^2)(xy+3z^2)\).

The condition gives \(a^3+b^3+c^3-3abc=0\). By the identity, this equals \((a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).

Since \(a+b+c>0\), the second factor is zero. But \(2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\). Hence \(a=b=c\).

We have \(x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2)\).

The first factor is \((x-y)^2+y^2\). For positive integers \(x,y\), it equals \(1\) only when \(y=1\) and \(x=y\), so \(x=y=1\).

Then the number is \(5\), which is prime. In all other cases both factors are greater than \(1\), so the number is composite. Answer: \((1,1)\).

\(x^6-1=(x^3-1)(x^3+1)\).

Also \(x^3-1=(x-1)(x^2+x+1)\), and \(x^3+1=(x+1)(x^2-x+1)\).

Thus \(x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1)\).

\(a^6-b^6=(a^3-b^3)(a^3+b^3)\).

Also \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Therefore \(a^2+ab+b^2\) is indeed a factor of \(a^6-b^6\).

Rewrite as \(x^2-2x-y^2=6\). Then \((x-1)^2-y^2=7\).

Thus \((x-1-y)(x-1+y)=7\). The possible factor pairs are \((1,7)\), \((7,1)\), \((-1,-7)\), \((-7,-1)\).

Solving the systems gives \((x,y)=(5,3),(5,-3),(-3,-3),(-3,3)\).

\(a^4+a^2b^2+b^4=(a^2+b^2)^2-a^2b^2\).

This is a difference of squares: \((a^2-ab+b^2)(a^2+ab+b^2)\).

Let the expression be \(E\). If \(a=b\), then \(E=0\), so \(a-b\) is a factor. Similarly, \(b-c\) and \(c-a\) are factors.

Since the degree is \(3\), \(E=k(a-b)(b-c)(c-a)\). Substitute \(a=1\), \(b=2\), \(c=3\): \(E=2\), and \((a-b)(b-c)(c-a)=2\). Hence \(k=1\).

Thus \(E=(a-b)(b-c)(c-a)\).

The expression is \((n+1)^4+4\). Put \(m=n+1\). Then \(m^4+4=(m^2-2m+2)(m^2+2m+2)\).

If \(|m|>1\), this is a non-trivial factorisation. If \(m=0\), the number is \(4\), not prime. If \(m=1\) or \(m=-1\), the number is \(5\), which is prime.

Hence \(n=0\) or \(n=-2\).

Assume \(n^4+4=m^2\). Then \(m^2-n^4=4\), so \((m-n^2)(m+n^2)=4\).

Both factors are positive. The possible pairs are \((1,4)\), \((2,2)\), \((4,1)\). The pair \((1,4)\) gives \(2m=5\), impossible. The pair \((2,2)\) gives \(n=0\), contradicting positivity. The pair \((4,1)\) is impossible by order.

Therefore \(n^4+4\) is not a square.

We have \((x-y)(x+y)=24\). Let \(u=x-y>0\), \(v=x+y\). Then \(uv=24\), and \(u\) and \(v\) have the same parity.

The suitable positive pairs are \((2,12)\) and \((4,6)\). From \(x=\frac{u+v}{2}\), \(y=\frac{v-u}{2}\), we get \((x,y)=(7,5)\), \((5,1)\).

If \(v<0\), then \(uv<0\) since \(u>0\), so there are no other solutions.

The left side vanishes when \(a=b\), when \(b=c\), and when \(c=a\). Hence it is divisible by \((a-b)(b-c)(c-a)\).

Since the degree is \(3\), the left side equals \(k(a-b)(b-c)(c-a)\). Substitute \(a=1\), \(b=2\), \(c=3\). The left side is \(-2\), while the product is \(2\). Thus \(k=-1\).

The identity is proved.

Let \(s_k=a^k+b^k+c^k\). Since \(a,b,c\) are roots of \(t^3+qt-r=0\), where \(q=ab+bc+ca\), \(r=abc\), each root satisfies \(t^3=-qt+r\).

Multiplying by \(t^2\) and summing over the three roots gives \(s_5=-q s_3+r s_2\).

When \(a+b+c=0\), we have \(s_3=3abc=3r\), and \(s_2=-2q\). Thus \(s_5=-3qr-2qr=-5qr\).

Since \(q=-\frac{s_2}{2}\), we obtain \(s_5=\frac{5}{2}rs_2=\frac{5}{2}abc(a^2+b^2+c^2)\).

Let the expression be \(E\). When \(a=b\), \(b=c\), or \(c=a\), it becomes zero, so \(E\) is divisible by \((a-b)(b-c)(c-a)\).

The remaining factor has degree \(2\) and is symmetric, so it has the form \(\alpha(a^2+b^2+c^2)+\beta(ab+bc+ca)\). Put \(c=0\). Then \(E=a^4b-b^4a=ab(a-b)(a^2+ab+b^2)\), while \((a-b)(b-c)(c-a)=-(ab)(a-b)\). Hence the remaining factor is \(-(a^2+ab+b^2)\) when \(c=0\).

Thus \(\alpha=-1\), \(\beta=-1\), and

\[E=-(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca).\]

The obvious attempt to take all factors large is not helpful: the product tends to decrease. The hidden idea is that small factors may become negative, and two negative factors give a positive product.

Let the original numbers be \(1,1,a\). The original product is \(a\), while the new product is \((-3)(-3)(a-4)=9a-36\).

We need the new product to be \(2020\) greater:

\[9a-36=a+2020.\]

Thus \(8a=2056\), so \(a=257\). Check: the original product is \(1\cdot1\cdot257=257\), the new product is \((-3)(-3)\cdot253=2277\), and \(2277-257=2020\). Therefore it is possible.

The point is not to guess all five numbers, but to create a large coefficient in front of one unknown. Let the numbers be \(1,1,1,1,a\). The original product is \(a\).

After decreasing each factor by \(4\), the product becomes \((-3)^4(a-4)=81a-324\).

We require

\[81a-324=78a.\]

Hence \(3a=324\), so \(a=108\). Check: the original product is \(108\), the new product is \((-3)^4\cdot104=8424\), and \(78\cdot108=8424\). Therefore \(1,1,1,1,108\) works.

The task is not about brute force; it is about controlling the sign. If we take five ones and one three, then after decreasing by \(4\) we get the factors \(-3,-3,-3,-3,-3,-1\), whose product is positive and equals \(243\).

Let the original numbers be \(1,1,1,1,1,3,a\). The original product is \(3a\), while the new product is \(243(a-4)\).

We want an \(80\)-fold increase:

\[243(a-4)=80\cdot3a.\]

Thus \(243a-972=240a\), so \(a=324\). Check: the original product is \(972\), the new product is \(243\cdot320=77760\), and \(80\cdot972=77760\). Therefore such a set exists.

\(x^2-9x+20=(x-4)(x-5)\). Hence \(x=4\) or \(x=5\).

Let \(u=x^2\). Then \(u^2-10u+9=0\), so \(u=1\) or \(u=9\).

Thus \(x=\pm1\) or \(x=\pm3\).

The numbers \(x,y\) are roots of \(t^2-8t+15=0\).

Thus \((t-3)(t-5)=0\). Therefore \((x,y)=(3,5)\) or \((5,3)\).

We get \(9-3a+12=0\), so \(21-3a=0\).

Therefore \(a=7\).

\(\left(x+\frac1x\right)^2=x^2+2+\frac1{x^2}=25\).

Hence \(x^2+\frac1{x^2}=23\).

From \(x^2+y^2=(x+y)^2-2xy\), we get \(25=49-2xy\), hence \(xy=12\).

Then \(x,y\) are roots of \(t^2-7t+12=0\), so \(t=3\) or \(t=4\). Answer: \((3,4)\), \((4,3)\).

We get \(u^2-13u+36=0\), so \((u-4)(u-9)=0\).

Since \(u=x^2\), we have \(x^2=4\) or \(x^2=9\). Hence \(x=\pm2,\pm3\).

Subtracting gives \(x^2-y^2+y-x=0\), so \((x-y)(x+y-1)=0\).

If \(x=y\), then \(x^2+x=12\), so \(x=3\) or \(x=-4\). This gives \((3,3)\), \((-4,-4)\).

If \(x+y=1\), then \(y=1-x\), and \(x^2+1-x=12\). Thus \(x^2-x-11=0\), so \(x=\frac{1\pm3\sqrt5}{2}\), \(y=\frac{1\mp3\sqrt5}{2}\).

\((x-y)(x+y)=21\). The two factors have the same parity, and the product is odd, so both are odd.

The pairs \((1,21)\), \((3,7)\), \((-1,-21)\), \((-3,-7)\) give \((11,10)\), \((5,2)\), \((-11,-10)\), \((-5,-2)\).

The discriminant is \(D=16-4a\). We need \(D=0\), so \(16-4a=0\).

Thus \(a=4\).

\(x^3+y^3=(x+y)^3-3xy(x+y)=216-18xy\).

From \(216-18xy=72\), we get \(xy=8\). Then \(x,y\) are roots of \(t^2-6t+8=0\), so \((x,y)=(2,4)\), \((4,2)\).

\((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\).

Substitute the data: \(36=12+2(xy+yz+zx)\). Hence \(xy+yz+zx=12\).

Let \(u=x+\frac1x=3\). Then \(u^3=x^3+\frac1{x^3}+3\left(x+\frac1x\right)\).

Thus \(27=x^3+\frac1{x^3}+9\), so \(x^3+\frac1{x^3}=18\).

Let the roots be \(m,n\). Then \(mn=12\), \(m+n=a\).

List integer divisor pairs: \((1,12)\), \((2,6)\), \((3,4)\), and the corresponding negative pairs. Thus \(a=13,8,7,-13,-8,-7\).

\(x^2-y^2=(x-y)(x+y)\). Since \(x+y=5\), we get \(5(x-y)=15\), so \(x-y=3\).

Solving \(x+y=5\), \(x-y=3\), gives \(x=4\), \(y=1\).

Let \(u=x^2-3x\). Then \(u^2-2u-8=0\), so \(u=4\) or \(u=-2\).

If \(x^2-3x=4\), then \(x^2-3x-4=0\), so \(x=4\) or \(x=-1\). If \(x^2-3x=-2\), then \(x^2-3x+2=0\), so \(x=1\) or \(x=2\).

\((x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3=3-6+3=0\).

A sum of squares is zero only when all squares are zero. Hence \(x=y=z=1\).

Subtract: \(x^2-y^2+2y-2x=0\), so \((x-y)(x+y-2)=0\).

If \(x=y\), then \(x^2+2x=9\), so \(x=-1\pm\sqrt{10}\). This gives two pairs \((-1+\sqrt{10},-1+\sqrt{10})\), \((-1-\sqrt{10},-1-\sqrt{10})\).

If \(x+y=2\), then \(y=2-x\). Substitution gives \(x^2+4-2x=9\), so \(x^2-2x-5=0\). Hence \(x=1\pm\sqrt6\), \(y=1\mp\sqrt6\).

We have \(xy-x-y=5\). Add \(1\): \((x-1)(y-1)=6\).

Since \(x,y\) are positive integers, \(x-1\) and \(y-1\) are positive divisors of \(6\). The pairs \((1,6)\), \((2,3)\), \((3,2)\), \((6,1)\) give \((x,y)=(2,7),(3,4),(4,3),(7,2)\).

Let \(p=xy\). Then \(x^2+y^2=4-2p\).

\(x^4+y^4=(x^2+y^2)^2-2x^2y^2=(4-2p)^2-2p^2=16-16p+2p^2\).

Since this equals \(2\), \(16-16p+2p^2=2\), so \(p^2-8p+7=0\). Thus \(p=1\) or \(p=7\). But real \(x,y\) with \(x+y=2\) require \(p\le1\). Hence \(p=1\), and \(x,y\) are roots of \(t^2-2t+1=0\). Therefore \(x=y=1\).

For \(x=3\): \(9-3(a+1)+a+6=0\), so \(12-2a=0\), hence \(a=6\).

For \(x=4\): \(16-4(a+1)+a+6=0\), so \(18-3a=0\), hence \(a=6\). Thus the only value is \(a=6\). For this value, both \(3\) and \(4\) are indeed roots.

Rewrite as \(x^2-2x+y^2-4y+5=0\).

This is \((x-1)^2+(y-2)^2=0\). A sum of squares is zero only when \(x=1\), \(y=2\).

The numbers \(x,y,z\) are roots of \(t^3-3t^2+3t-1\).

But \(t^3-3t^2+3t-1=(t-1)^3\). Therefore all three roots equal \(1\), so \(x=y=z=1\).

Without loss of generality, let \(x\le y\le z\). If \(x\ge3\), then \(xyz\ge3yz\), while the right side is at most \(3z+2\); for \(y\ge3\), \(z\ge3\), this is impossible. Hence \(x=1\) or \(x=2\).

If \(x=1\), then \(yz=y+z+3\), so \((y-1)(z-1)=4\). This gives \((y,z)=(2,5),(3,3),(5,2)\).

If \(x=2\), then \(2yz=y+z+4\). For \(y=1\), \(z=5\); for \(y=2\), \(z=2\); for \(y\ge3\), the left side is too large: \(2yz-y-z-4\ge6z-3-z-4=5z-7>0\). Taking permutations into account, all solutions are permutations of \((1,2,5)\), \((1,3,3)\), and \((2,2,2)\).

Let \(s=x+y\), \(p=xy\). Then \(s+p=11\), and \(x^2+y^2=s^2-2p=25\).

Since \(p=11-s\), we get \(s^2-2(11-s)=25\), so \(s^2+2s-47=0\). Thus \(s=-1\pm4\sqrt3\), \(p=11-s\).

Real \(x,y\) require \(s^2-4p\ge0\), which holds for both values. Then \(x,y\) are roots of \(t^2-st+p=0\). Thus for each \(s=-1\pm4\sqrt3\), \(p=11-s\), the pairs \((x,y)\) are the two permutations of the roots of \(t^2-st+p=0\).

Answer: no, not necessarily. We show how to find a counterexample.

From the equality of the first two expressions, \(x+2y^2=y+2x^2\). Move all terms to one side:

\[x-y-2(x^2-y^2)=0.\]

Thus \((x-y)(1-2(x+y))=0\). Therefore for each pair of variables, either they are equal or their sum is \(\frac12\).

Take \(x=y=\frac16\), \(z=\frac13\). Then \(x+y=\frac13\), but \(x+z=y+z=\frac12\). Check:

\[x+2y^2+2z^2=\frac16+\frac1{18}+\frac2{9}=\frac49,\]

and the other two expressions are also equal to \(\frac49\). However, \(x,y,z\) are not all equal. Hence the statement is false.

Since \(P(1)=0\), \(P(x)\) is divisible by \(x-1\). Division gives \(P(x)=(x-1)(x^2-x-6)=(x-1)(x-3)(x+2)\).

\(P(-1)=-1+2-7=-6\). The remainder is \(-6\).

\((x+4)(x-6)=x^2-2x-24\). Hence \(a=-2\), \(b=-24\).

The polynomial \(R=P-Q\) has degree at most \(2\) and three distinct roots: \(-1,0,2\). Hence \(R\equiv0\), so \(P=Q\).

For a monomial \(cx^k\), the difference of values is \(c(17^k-5^k)\), which is divisible by \(12\). The sum of such differences is also divisible by \(12\).

The value at \(x=2\) is \(8+4a-8-4a=0\). Therefore the divisibility holds for every \(a\).

Let the common root be \(r\). Then \(r^2+ar+b=0\) and \(r^2+br+a=0\). Subtracting gives \((a-b)(r-1)=0\). Since \(a\ne b\), \(r=1\).

The discriminant equals \((u+v)^2-4uv=u^2-2uv+v^2=(u-v)^2\).

If \(P\) were non-zero, it would have at most \(4\) roots. But it has at least \(5\) roots. Contradiction. Hence \(P\equiv0\).

For an integer-coefficient polynomial, \(P(8)-P(2)\) is divisible by \(6\). But the condition gives \(P(8)-P(2)=-2\), which is not divisible by \(6\). Contradiction.

The condition gives \(1000^2b+1000c+a=1000^2c+1000a+b\). Moving terms and dividing by \(999\), we get \(1001b-1000c-a=0\), so \(a=1001b-1000c\).

The value of the first trinomial at \(1\) is \(a+b+c=1002b-999c=3(334b-333c)\). It is always divisible by \(3\). The number \(2024\) is not divisible by \(3\), so this is impossible.

Let \(r\) be a common root. Then \(P(r)=0\), so \(P(P(P(r)))=P(P(0))\). By the condition this is \(0\), hence \(P(P(0))=0\).

Write \(P(x)=x^2+ux+v\). Then \(P(0)=v\), and \(P(v)=0\) gives \(v^2+uv+v=v(v+u+1)=0\).

If \(v=0\), then \(P(0)=0\). If \(v+u+1=0\), then \(P(1)=1+u+v=0\). In both cases \(P(0)P(1)=0\).

Since the product has exactly three distinct roots, the trinomials have a common root \(r\). Also \(u\ne v\), otherwise there would be only two roots.

Subtract: \(0=(r^2+ur+v)-(r^2+vr+u)=(u-v)(r-1)\). Hence \(r=1\). Substitution gives \(1+u+v=0\).

Let the other roots be \(r_1,r_2\). By Vieta, \(1+r_1=-u\), \(1+r_2=-v\). Therefore the sum of the three distinct roots is \(1+r_1+r_2=1-u-v-2=-u-v-1=0\).

Let \(P(x)=(x-a_1)(x-a_2)\), \(Q(x)=(x-b_1)(x-b_2)\). The condition says \(Q(a_1)+Q(a_2)=P(b_1)+P(b_2)\).

Expanding only the needed sums gives \((a_1-b_1)(a_1-b_2)+(a_2-b_1)(a_2-b_2)=(b_1-a_1)(b_1-a_2)+(b_2-a_1)(b_2-a_2)\).

After cancellation this is equivalent to \((a_1-a_2)^2=(b_1-b_2)^2\). For a monic quadratic, the discriminant is the square of the difference of roots. Hence the discriminants are equal.

If \(k<5\), then from any suitable polynomial \(P\) we can build another one: \(Q(x)=P(x)+(x-n_1)\cdots(x-n_k)\). It remains monic of degree \(5\), but \(Q(n_i)=P(n_i)\). Thus uniqueness is impossible.

For \(k=5\), choose \(n_i=10i\) and announce that the product of values is \(1\). The polynomial \(P(x)=\prod_{i=1}^5(x-n_i)+1\) works.

If another monic integer polynomial \(Q\) also gives product \(1\), then all \(Q(n_i)=\pm1\). But \(Q(n_i)-Q(n_j)\) is divisible by \(n_i-n_j\), while \(2\) is not divisible by \(10\) or more. Hence all values \(Q(n_i)\) are equal; since there are five of them and their product is \(1\), all equal \(1\). Then \(P-Q\) has five roots and degree at most \(4\), so \(P=Q\).

The polynomial \(h(x)=g(7-x)\) is also monic quadratic. The conditions give \(h(2)=g(5)=f(2)\) and \(h(5)=g(2)=f(5)\).

The difference \(h-f\) has degree at most \(1\), but has two roots \(2\) and \(5\). Hence \(h=f\).

If \(r\) is a root of \(f\), then \(7-r\) is a root of \(g\). The roots form two pairs with sum \(7\), so the total sum is \(14\).

The discriminants are \(D_1=p^2-4q\), \(D_2=p^2-4q-4\), \(D_3=p^2-4q-8\).

Since the first two trinomials have integer roots, \(D_1=m^2\), \(D_2=n^2\) for non-negative integers \(m,n\). Thus \(m^2-n^2=4\), so \((m-n)(m+n)=4\). The two factors have the same parity, so only \(m-n=m+n=2\) is possible, giving \(n=0\).

Then \(D_3=D_2-4=-4<0\). Therefore the third trinomial has no real roots.

By Vieta for integer roots \(r,s\): \(-b/a=r+s\), \(c/a=rs\). Hence \(a\mid b\) and \(a\mid c\).

Suppose the numbers used are \(k+1,\ldots,k+3n\). Order the leading coefficients: \(a_1<\cdots

For the trinomial with leading coefficient \(a_n\), the coefficients \(b,c\) must be positive multiples of \(a_n\), so one of them is at least \(3a_n\). But every coefficient is at most \(k+3n\). Thus \(3(k+n)\le k+3n\), hence \(k=0\). Then \(a_i=i\).

If \(i\) is even, all three coefficients in that trinomial are even. If \(i\) is odd, at least one coefficient in the triple must be even; otherwise the trinomial has odd value at every integer \(x\) and cannot have an integer root. Thus the number of even coefficients is at least \(n+2\lfloor n/2\rfloor\), which for \(n>10\) exceeds the number of even integers among \(1,\ldots,3n\). Contradiction.

The axis of the parabola passes through the midpoint of the roots, \(c=\frac{a+b}{2}\), and the vertex has coordinates \((c,-1)\). Hence \(y=\lambda(x-c)^2-1\). Substituting \(x=a\) gives \(0=\lambda\left(\frac{b-a}{2}\right)^2-1\), so \(\lambda=\frac{4}{(b-a)^2}\).

For two parabolas, common points satisfy \(\frac{4}{w_1^2}(x-c_1)^2=\frac{4}{w_2^2}(x-c_2)^2\). This is equivalent to two linear equations \(\frac{x-c_1}{w_1}=\pm\frac{x-c_2}{w_2}\). If \(w_1\ne w_2\) and \(c_1\ne c_2\), both give one distinct solution. If the width or the axis coincides, one solution merges or disappears, leaving exactly one common point.

Let \(P(a)-P(b)=N^2\). For \(P(x)=ux^2+vx+w\), \(P(a)-P(b)=(a-b)(u(a+b)+v)\).

Take \(c=a+k\), \(d=b-k\). Then \(c+d=a+b\), and \(c-d=a-b+2k\), so \(P(c)-P(d)=(a-b+2k)(u(a+b)+v)\).

If \(k=(a-b)m\), then \(P(c)-P(d)=(2m+1)N^2\). Choose \(m\) so that \(2m+1\) is a square of an odd number: \(2m+1=(2t+1)^2\). For \(t=1,2,\ldots,1001\), we get more than \(1000\) different pairs.

The polynomial \((x-a_1)(x-a_2)(x-a_3)-t\) has roots \(c_1,c_2,c_3\), so it equals \((x-c_1)(x-c_2)(x-c_3)\).

Substitute \(-x\): \((-x-a_1)(-x-a_2)(-x-a_3)-t=(-x-c_1)(-x-c_2)(-x-c_3)\).

Multiplying by \(-1\), we obtain \((x+a_1)(x+a_2)(x+a_3)+t=(x+c_1)(x+c_2)(x+c_3)\). Therefore at \(x=-a_1,-a_2,-a_3\), the left side of the equation equals \(t\). Answer: \(-a_1,-a_2,-a_3\).

The tangent to \(y=x^2\) at \(u\) is \(y=2ux-u^2\), and at \(v\) it is \(y=2vx-v^2\). Their intersection is \(x=\frac{u+v}{2}\), \(y=uv\).

Since the graphs intersect at \(u,v\), the polynomial \((p-1)x^2+qx+r\) has roots \(u,v\). Hence \((p-1)x^2+qx+r=(p-1)(x-u)(x-v)\).

Substitute \(C\) into \(y=px^2+qx+r\). Using \(q=-(p-1)(u+v)\), \(r=(p-1)uv\), we get \(uv=p\left(\frac{u+v}{2}\right)^2-(p-1)(u+v)\frac{u+v}{2}+(p-1)uv\). Simplification gives \((p-2)(u-v)^2=0\). Since \(u\ne v\), \(p=2\).

Let \(P(n)=an^2+bn+c\), \(Q(n)=un^2+vn+w\). If \(P(k)\ne P(m)\), the abscissa of the intersection of \(L_k,L_m\) is

\[x_{km}=\frac{Q(m)-Q(k)}{P(k)-P(m)}=-\frac{u(k+m)+v}{a(k+m)+b}.\]

If three lines \(L_k,L_m,L_s\) are concurrent, then \(x_{km}=x_{ks}\). Since \(m\ne s\), equality of the fractions implies \(ub-av=0\), so the expression for \(x_{km}\) does not depend on \(k,m\). Thus all pairs meet at the same abscissa \(x_0\).

Substituting \(x_0\), the values \(P(n)x_0+Q(n)\) are equal for three different \(n\). This is a quadratic polynomial in \(n\); after subtracting the common value, it has three roots, so it is identically zero. Therefore all \(L_n\) pass through one point.

By Vieta, \(r_1+r_2=-\frac ba\). Therefore

\[f(r_1+r_2)=f\left(-\frac ba\right)=a\frac{b^2}{a^2}-\frac{b^2}{a}+c=c.\]

Hence \(c=137\).

\(a_{15}=4+14\cdot5=74\).

\(b_6=2\cdot3^5=486\).

\(\frac1{k(k+1)}=\frac1k-\frac1{k+1}\). The sum is \(1-\frac1{11}=\frac{10}{11}\).

By induction, \(a_n=n^2\). Hence \(a_{20}=400\).

The general term is \(a_n=53-5(n-1)\). Positivity gives \(58-5n>0\), so \(n\le11\). The last term is \(a_{11}=3\).

The sum is \(\frac{11(53+3)}2=308\).

We need \(3c=4+c\), so \(c=2\). Then \(a_{n+1}+2=3(a_n+2)\).

\(a_1+2=4\), hence \(a_n+2=4\cdot3^{n-1}\), and \(a_n=4\cdot3^{n-1}-2\).

\(F_{n+3}=F_{n+2}+F_{n+1}=F_n+2F_{n+1}\). Since \(F_{n+1}\ge F_n\) for \(n\ge1\), \(F_{n+3}\ge3F_n\), hence \(F_{n+3}\ge2F_n\).

The differences are \(2n+1\) from \(a_n\) to \(a_{n+1}\). This corresponds to \(a_n=n^2+1\). Check: \(1^2+1=2\), \(2^2+1=5\), and so on.

\(\frac{1}{(k+1)(k+3)}=\frac12\left(\frac1{k+1}-\frac1{k+3}\right)\).

The sum is \(\frac12\left(\frac12+\frac13-\frac1{n+2}-\frac1{n+3}\right)\).

Then \(B>A>1\), \(x_{n+1}=2^{n+1}(B-A)\), and \(x_n=2^n(B^2-A^2)=2^{n+1}(B-A)\frac{A+B}{2}\).

Since \(\frac{A+B}{2}>1\), we get \(x_n>x_{n+1}\).

Base case is clear: \(4\mid a_1\). If \(4\mid a_n\), then \(5a_n+4\) is also divisible by \(4\). Hence \(4\mid a_{n+1}\).

\(\frac1{a_{n+1}}=\frac{a_n+2}{2a_n}=\frac12+\frac1{a_n}\). Thus \(\frac1{a_n}=\frac12+\frac{n-1}{2}=\frac n2\). Hence \(a_n=\frac2n\).

The difference sequence has \(n-1\) terms, each equal to \(-1,0,1\), giving \(3^{n-1}\) choices.

For each difference sequence, all original sequences are obtained by adding the same constant to all terms. Exactly one has minimum \(1,2,3,4\). Therefore the answer is \(4\cdot3^{n-1}\).

Base \(n=2\): \(F_3F_1-F_2^2=2\cdot1-1=1=(-1)^2\).

If \(F_{n+1}F_{n-1}-F_n^2=(-1)^n\), then \(F_{n+2}F_n-F_{n+1}^2=(F_{n+1}+F_n)F_n-F_{n+1}^2=-(F_{n+1}F_{n-1}-F_n^2)=(-1)^{n+1}\).

\(1-a_{n+1}=1-2a_n+a_n^2=(1-a_n)^2\). If \(0

Also \(a_{n+1}-a_n=a_n(1-a_n)>0\). Therefore the sequence is increasing.

For \(k\ge2\), \(k^2>k(k-1)\), so \(\frac1{k^2}<\frac1{k(k-1)}=\frac1{k-1}-\frac1k\).

Thus the sum is less than \(1+\sum_{k=2}^{n}\left(\frac1{k-1}-\frac1k\right)=1+1-\frac1n<2\).

We prove the stronger statement that \(a_n\) is divisible by the product of all previous terms. The base is clear.

If \(a_n\) is divisible by \(a_1\cdots a_{n-1}\), then \(a_{n+1}=a_n(a_n+1)\) is divisible by \(a_1\cdots a_{n-1}\cdot a_n\). Hence the claim follows by induction.

Let \(a_n=3n^2+bn+c\). From \(a_1=2\): \(3+b+c=2\). From \(a_2=9\): \(12+2b+c=9\).

Subtracting gives \(9+b=7\), so \(b=-2\). Then \(c=1\). Answer: \(a_n=3n^2-2n+1\).

Construction: draw \(A_2A_4,A_3A_5,\ldots,A_{n-2}A_n\), then \(A_1A_3,A_1A_4,\ldots,A_1A_{n-1}\). There are \((n-3)+(n-3)=2n-6\) diagonals, and the order can be chosen so that each new one intersects at most one old one.

For the upper bound, use induction on \(n\). The triangle is clear. Let the last diagonal be \(A_1A_k\). It intersects at most one previous diagonal \(d\). All other diagonals lie inside one of the two polygons \(A_1A_2\ldots A_k\) or \(A_kA_{k+1}\ldots A_nA_1\). By induction, their number is at most \((2k-6)+(2(n+2-k)-6)=2n-8\). Adding the last diagonal and possibly \(d\), we get at most \(2n-6\).

For \(n=1\), \(28t\in\mathbb Z\); for \(n=2\), \(80t\in\mathbb Z\); for \(n=3\), \(162t\in\mathbb Z\). The greatest common divisor of \(28,80,162\) is \(2\), so \(2t\in\mathbb Z\).

Conversely, \(n(n+3)(n+6)\) is always even: if \(n\) is even this is clear, and if \(n\) is odd then \(n+3\) is even. Thus every \(t=\frac{k}{2}\), \(k\in\mathbb Z\), works. Answer: \(t=\frac{k}{2}\).

Let the given number be \(n(n+1)\). Then

\[n(n+1)=\frac{n(n+2)(n+1)^2}{(n+1)(n+2)}=\frac{(n^2+2n)(n^2+2n+1)}{(n+1)(n+2)}.\]

The numerator is the product of consecutive integers \(n^2+2n\) and \(n^2+2n+1\), and the denominator \((n+1)(n+2)\) is also rectangular. Thus the representation is found.

Let \(u=(a,b-a)\), \(v=(b-a,-a)\). Then \(a u+(b-a)v=(a^2+(b-a)^2,0)=(D,0)\).

Also \((b-a)u-a v=(0,D)\). Thus the marking is periodic under shifts \((D,0)\) and \((0,D)\).

In a fundamental \(D\times D\) square, the number of marked cells equals the area of the square divided by the area of the fundamental parallelogram of the lattice. The determinant of \(u,v\) is \(-a^2-(b-a)^2=-D\), so that area is \(D\). Hence a square of area \(D^2\) contains exactly \(D\) marked cells.

Fix the differences \(b_i=a_{i+1}-a_i\). Each difference can be \(-2,-1,0,1,2\), giving \(5^{n-1}\) choices.

For each difference sequence, exactly one original sequence has minimum \(1,2,3,4,5\). Thus there are \(5^n\) sequences with minimum at most \(5\).

The extra ones are those with no \(4\) or \(5\). If such a sequence has minimum at most \(5\), step size at most \(2\), and contains a value greater than \(5\) and a value less than \(4\), it must pass through \(4\) or \(5\). Hence the extra sequences consist only of \(1,2,3\). There are \(3^n\) of them. Answer: \(5^n-3^n\).

Then \(B>A>1\), \(y_{n+1}=3^{n+1}(B-A)\), and \(y_n=3^n(B^3-A^3)=3^{n+1}(B-A)\frac{B^2+AB+A^2}{3}\).

Since \(A,B>1\), the average \(\frac{B^2+AB+A^2}{3}>1\). Therefore \(y_n>y_{n+1}\).

Take \(a=n\), \(d=n+5\), \(b=n+1\), \(c=n+4\), \(e=n+2\), \(f=n+3\).

Then \(b+c=e+f=2n+5\), and \(a+d=2n+5\). Hence \(\frac a{b+c}+\frac d{e+f}=\frac{a+d}{2n+5}=1\).

Among the given numbers are \(n,n+1,\ldots,n+7\). Take the first six: \(n,\ldots,n+5\).

As in the previous construction, set \(a=n\), \(d=n+5\), \(b=n+1\), \(c=n+4\), \(e=n+2\), \(f=n+3\). Then both denominators equal \(2n+5\), and \(a+d=2n+5\). Hence the sum is \(1\).

We have \(x^2-8x=(x-4)^2-16\) and \(y^2+2y=(y+1)^2-1\). Hence the expression equals \((x-4)^2+(y+1)^2+3\).

A sum of squares is nonnegative, so the minimum is \(3\), attained at \(x=4\), \(y=-1\).

From \((a+b+c)^2=0\) we get \(a^2+b^2+c^2+2(ab+bc+ca)=0\), which gives the first equality.

Also, \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\). The right-hand side is zero, hence \(a^3+b^3+c^3=3abc\).

\(x^2+y^2=(x+y)^2-2xy=49-20=29\).

Next, \(x^3+y^3=(x+y)^3-3xy(x+y)=343-210=133\).

\(\frac{(x+y)^2}{xy}=\frac{x^2+2xy+y^2}{xy}=\frac{x}{y}+2+\frac{y}{x}=3+2=5\).

Since the expression is homogeneous, the factor \(t\) cancels. We get \(\frac{4t^2+9t^2+25t^2}{6t^2+15t^2+10t^2}=\frac{38}{31}\).

We get \(s+p=7\) and \(s^2-2p=10\). Since \(p=7-s\), it follows that \(s^2+2s-24=0\), so \(s=4\) or \(s=-6\).

If \(s=4\), then \(p=3\), and \(x,y\) are the roots of \(t^2-4t+3=0\), hence \(\{x,y\}=\{1,3\}\). If \(s=-6\), then \(p=13\), but the discriminant \(s^2-4p=36-52<0\), so no real \(x,y\) exist.

Answer: \((1,3)\) and \((3,1)\).

From \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\) we get \(36=a^2+b^2+c^2+22\), so \(a^2+b^2+c^2=14\).

Then \((a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(14-11)=6\).

From \((p+q+r)^2=0\) we get \(18+2(pq+qr+rp)=0\), so \(pq+qr+rp=-9\).

Therefore \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{pq+qr+rp}{pqr}=\frac{-9}{6}=-\frac{3}{2}\).

The left-hand side equals \((x-2)^2+4(y-1)^2+5\). This is always at least \(5\), so it cannot be zero.

After dividing numerator and denominator by \(b\), we get \(\frac{t-1}{t+1}=\frac{1}{3}\). Thus \(3t-3=t+1\), so \(2t=4\) and \(t=2\).

The left-hand side equals \(2(a^2+b^2+c^2-ab-bc-ca)\). Since \(a+b+c=0\), we have \(ab+bc+ca=-\frac{a^2+b^2+c^2}{2}\).

Substitution gives \(2\left(a^2+b^2+c^2+\frac{a^2+b^2+c^2}{2}\right)=3(a^2+b^2+c^2)\).

From the conditions we get \(xy+yz+zx=-1\). Hence \(x,y,z\) are roots of the polynomial \(t^3+(xy+yz+zx)t-xyz=t^3-t-xyz\).

Therefore each of \(x,y,z\) satisfies \(t^3-t=xyz\). Thus \(x^3-x=y^3-y=z^3-z=xyz\).

Let \(s=x+y=3\), \(p=xy\). Then \(x^2+y^2+xy=s^2-p\). We get \(9-p=7\), hence \(p=2\).

The numbers \(x,y\) are roots of \(t^2-3t+2=0\), so \(\{x,y\}=\{1,2\}\). Answer: \((1,2)\), \((2,1)\).

Since \(x+y+z=0\), we have \(y+z=-x\), \(z+x=-y\), \(x+y=-z\). Therefore the sum is \(\frac{x}{-x}+\frac{y}{-y}+\frac{z}{-z}=-1-1-1=-3\).

\(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)\). Hence \(27=x^3+\frac{1}{x^3}+9\), so the desired value is \(18\).

We have \((a-1)^2+(b-1)^2+(c-1)^2=a^2+b^2+c^2-2(a+b+c)+3=3-6+3=0\).

A sum of squares is zero only when each square is zero. Hence \(a=b=c=1\).

From the zero sum we get \(xy+yz+zx=-3\). Then \((xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)\), and the second term is zero. Hence \(x^2y^2+y^2z^2+z^2x^2=9\).

Now \((x^2+y^2+z^2)^2=x^4+y^4+z^4+2(x^2y^2+y^2z^2+z^2x^2)\). Therefore \(36=x^4+y^4+z^4+18\), so \(x^4+y^4+z^4=18\).

From the first two conditions, \(ab+bc+ca=-3\). From \(a^3+b^3+c^3=3abc\) we get \(abc=2\).

Thus \(a,b,c\) are roots of \(t^3-3t-2=0\). The factorization \(t^3-3t-2=(t-2)(t+1)^2\) gives the roots \(2,-1,-1\). Hence all solutions are permutations of \((2,-1,-1)\).

Since \(\frac{x^2+y^2}{xy}=\frac{s^2-2p}{p}=\frac{5}{2}\), we have \(\frac{s^2}{p}=\frac{9}{2}\).

Then \(\left(\frac{x-y}{x+y}\right)^2=\frac{s^2-4p}{s^2}=1-\frac{4p}{s^2}=1-\frac{8}{9}=\frac{1}{9}\). Thus the possible values are \(\frac{1}{3}\) and \(-\frac{1}{3}\). Both are attained, for example, by the ratios \(x:y=2:1\) and \(1:2\).

From the conditions we get \(ab+bc+ca=-1\). Let \(q=abc\). Then \(a,b,c\) are three real roots of \(F(t)=t^3-t-q\).

If the cubic \(t^3-t-q\) has three real roots, then \(q\) must lie between the local maximum and local minimum values of the function \(t^3-t\). These extrema occur at \(3t^2-1=0\), so \(t=\pm\frac{1}{\sqrt{3}}\).

We get \(\left(\frac{1}{\sqrt{3}}\right)^3-\frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}\) and \(\left(-\frac{1}{\sqrt{3}}\right)^3+\frac{1}{\sqrt{3}}=\frac{2}{3\sqrt{3}}\). Therefore \(-\frac{2}{3\sqrt{3}}\le q\le\frac{2}{3\sqrt{3}}\), as required.

Answer: no.

Since the graphs are lines and the coefficients \(v_i\) are distinct, any two graphs intersect at most once. Consider the order of the graphs by height for a very large \(t\), after all positive intersections.

The graph \(f_1\) was the lowest at \(t=0\). Each of its meetings moves it one position upward in the current order. If it has exactly four meetings, then finally it is fifth from the bottom.

Similarly, \(f_9\) was the highest at \(t=0\). Each of its meetings moves it one position downward. If it has exactly four meetings, then finally it is also fifth from the bottom. Two different graphs cannot occupy the same position. Contradiction.

Take any two given numbers \(x\) and \(y\). Consider two choices of four numbers that differ only in this: in one choice \(x\) is in the square group and \(y\) is in the other group, while in the second choice they are swapped. The other five contributions are the same.

By the condition, the sums are equal, so \(x^2+(2y-y^2)=y^2+(2x-x^2)\). Moving everything to one side gives \(2x^2-2y^2-2x+2y=0\), that is \((x-y)(x+y-1)=0\).

Thus any two numbers in the set are either equal or have sum \(1\). Fix one number \(x\). Then every other number is either \(x\) or \(1-x\). Hence there are at most two distinct values among the seven numbers. By the pigeonhole principle, one of them occurs at least four times.

Since \(S\) and \(C\) are rational, \(S^2+C^2\) is rational. Expanding, \[S^2+C^2=2+2(\sin48t\sin49t+\cos48t\cos49t)=2+2\cos t.\] Hence \(\cos t\) is rational.

For every positive integer \(n\), \(\cos nt\) is a polynomial with integer coefficients in \(\cos t\). Therefore \(\cos48t\) is rational. Since \(C=\cos48t+\cos49t\) is rational, \(\cos49t=C-\cos48t\) is rational as well.

We have \(\cos y=A-\sin x\) and \(\sin y=\cos x-B\). Square and add: \[(A-\sin x)^2+(\cos x-B)^2=1.\]

After expanding and using \(\sin^2x+\cos^2x=1\), we get \(A^2+B^2-2A\sin x-2B\cos x=0\), that is \(2A\sin x+2B\cos x=A^2+B^2\).

Let \(N\) be a common denominator of the rational numbers \(A\) and \(B\). Then \(m=2AN^2\) and \(n=2BN^2\) are positive integers, and \((A^2+B^2)N^2\) is a positive integer. Multiplying the equality by \(N^2\) gives the result.

Answer: no.

For neighboring numbers \(u\neq v\), \[\left(\frac{uv}{u-v}\right)^{-1}=\frac{u-v}{uv}=\frac{1}{v}-\frac{1}{u}.\] Hence \(\frac{uv}{u-v}\) is rational if and only if the difference \(\frac{1}{v}-\frac{1}{u}\) is rational.

Assume that all values are rational except possibly the value for the pair \(a_{2028},a_1\). Then the differences \(\frac{1}{a_2}-\frac{1}{a_1}\), \(\frac{1}{a_3}-\frac{1}{a_2}\), ..., \(\frac{1}{a_{2028}}-\frac{1}{a_{2027}}\) are rational. Their sum is \(\frac{1}{a_{2028}}-\frac{1}{a_1}\), so it is rational as well.

Therefore \(\frac{a_{2028}a_1}{a_{2028}-a_1}\) is rational. This contradicts the assumption that exactly one value is irrational.

We have \((a-b)^2=a^2-2ab+b^2\ge0\). Hence \(a^2+b^2\ge2ab\). Equality occurs when \(a=b\).

By AM-GM, \(x+\frac{1}{x}\ge2\sqrt{x\cdot\frac{1}{x}}=2\). Equality holds at \(x=1\), so the minimum is \(2\).

\(2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\ge0\). Therefore the inequality holds. Equality occurs when \(a=b=c\).

\(\frac{x+y}{2}\ge\sqrt{xy}\), so \(5\ge\sqrt{xy}\). Hence \(xy\le25\). Equality occurs when \(x=y=5\).

By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). The left side is \(2\), so \(\sqrt[3]{abc}\le2\), hence \(abc\le8\). Equality occurs when \(a=b=c=2\).

By AM-GM, \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\). Equality holds when \(a=b\).

By Cauchy, \(\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{(x+y)^2}{x+y}=x+y\). Equality holds when \(x=y\).

By Cauchy-Schwarz in Engel form, \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b}\). Equality occurs when \(\frac{x}{a}=\frac{y}{b}\).

From the three-squares inequality, \((a+b+c)^2\ge3(ab+bc+ca)\). Since \(a+b+c=1\), we get \(1\ge3(ab+bc+ca)\), so \(ab+bc+ca\le\frac{1}{3}\). Equality holds at \(a=b=c=\frac{1}{3}\).

By AM-GM, \(\frac{x+y+z}{3}\ge\sqrt[3]{xyz}=1\). Therefore \(x+y+z\ge3\). Equality occurs when \(x=y=z=1\).

By Cauchy, the sum is at least \(\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(a+b+c)^2}{2(ab+bc+ca)}\).

Since \((a+b+c)^2\ge3(ab+bc+ca)\), we get \(\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac{3}{2}\). Equality holds when \(a=b=c\).

By Cauchy, \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\ge(1+1+1)^2=9\). Since \(x+y+z=1\), the inequality follows. Equality holds when \(x=y=z=\frac{1}{3}\).

By Cauchy, \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{(a+b+c)^2}{a+b+c}=a+b+c\). Equality holds when \(a=b=c\).

By AM-GM, \(1+x\ge2\sqrt{x}\), \(1+y\ge2\sqrt{y}\), \(1+z\ge2\sqrt{z}\). Multiplying, \((1+x)(1+y)(1+z)\ge8\sqrt{xyz}=8\). Equality holds when \(x=y=z=1\).

The first part is obvious because a sum of squares is nonnegative; equality is possible only when all squares are zero.

For the second part, set \(m=\frac{x+y+z}{3}\), \(a=x-m\), \(b=y-m\), \(c=z-m\). Then \(a+b+c=0\), so \((x-m)^2+(y-m)^2+(z-m)^2\ge0\). Expanding gives \(x^2+y^2+z^2\ge3m^2=\frac{(x+y+z)^2}{3}\).

The difference is \(ax+cz-az-cx=(c-a)(z-x)\). Since \(c-a\ge0\) and \(z-x\ge0\), it is nonnegative. Therefore \(ax+by+cz\ge az+by+cx\).

By Cauchy, the left-hand side is at least \(\frac{(x+y+z)^2}{(y+z)+(z+x)+(x+y)}=\frac{(x+y+z)^2}{2(x+y+z)}=\frac{x+y+z}{2}\). Equality holds when \(x=y=z\).

By Cauchy, the left-hand side is at least \(\frac{(1+1+1)^2}{(3+a)+(3+b)+(3+c)}=\frac{9}{9+a+b+c}=\frac{9}{12}=\frac{3}{4}\). Equality holds when \(a=b=c=1\).

Since \(a+b+c=1\), we have \(1-a=b+c\), \(1-b=c+a\), \(1-c=a+b\). Thus the inequality becomes \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\), which is Nesbitt's inequality.

By Cauchy, the left-hand side is at least \(\frac{(x+y+z)^2}{(x+1)+(y+1)+(z+1)}=\frac{36}{9}=4\). Equality holds when \(x=y=z=2\).

By Cauchy, \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge(\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}}+\sqrt{c}\cdot\frac{1}{\sqrt{c}})^2=9\). Equality holds when \(a=b=c\).

By Cauchy, the left-hand side is at least \(\frac{(a+b+c)^2}{(a+b)+(b+c)+(c+a)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}\).