#19 Mixed Problems I

The numbers \(n\) and \(n+1\) have opposite parity, so one is divisible by \(2\). Hence their product is divisible by \(2\).

If \(d\mid n\) and \(d\mid n+1\), then \(d\mid (n+1)-n=1\). Thus \(d=1\), so the GCD is \(1\).

If \(n\) is even, then \(n^2\equiv0\pmod4\). If \(n\) is odd, then \(n\equiv1\) or \(3\pmod4\), and in both cases \(n^2\equiv1\pmod4\). Only \(0\) and \(1\) are possible.

We have \(x^2-y^2=(x-y)(x+y)\). This is useful when an equation fixes a difference of squares: then the problem reduces to factor pairs.

The last digits are \(7,9,3,1\). Since \(2025\equiv1\pmod4\), the last digit is \(7\).

We have \(n^3-n=n(n-1)(n+1)\). Among three consecutive integers, one is divisible by \(3\), so the product is divisible by \(3\).

Since \(n^2-1=(n-1)(n+1)\), the number \(n+1\) divides \(n^2-1\). Therefore \(\gcd(n^2-1,n+1)=n+1\).

We get \((x+1)(y+1)=24\). Factor pairs of \(24\) greater than \(1\) are \((2,12),(3,8),(4,6),(6,4),(8,3),(12,2)\). Hence \((x,y)=(1,11),(2,7),(3,5),(5,3),(7,2),(11,1)\).

The left side modulo \(4\) can only be \(0,1,2\). The right side is \(3\pmod4\). Contradiction.

Let \(n=3k+1\). Then \(3k+1\equiv2\pmod5\), so \(3k\equiv1\pmod5\), \(k\equiv2\pmod5\). Hence \(n=3(5t+2)+1=15t+7\). Answer: \(n\equiv7\pmod{15}\).

Since \(10 ot\equiv1\pmod{11}\), but \(10^2\equiv1\pmod{11}\), the order of \(10\) modulo \(11\) is \(2\). Thus the period is \(2\).

If \(d\mid n\), it pairs with \(\frac{n}{d}\). The two divisors are distinct except when \(d=\frac{n}{d}\), i.e. \(d^2=n\). Thus an unpaired divisor appears exactly for squares, and only then is the number of divisors odd.

For \(n=0,1,2,3,4,5,6\), the expression \(n^2+n+1\) gives residues \(1,3,0,6,0,3,1\). Hence \(n\equiv2\) or \(4\pmod7\).

Modulo \(n+2\), we have \(n\equiv-2\), so \(n^2+1\equiv4+1=5\). Thus the GCD is \(\gcd(n+2,5)\). It is \(5\) if \(n\equiv3\pmod5\), and \(1\) otherwise.

We get \(xy-3x-2y=0\). Add \(6\): \((x-2)(y-3)=6\). Positive factor pairs of \(6\) are \((1,6),(2,3),(3,2),(6,1)\). Thus \((x,y)=(3,9),(4,6),(5,5),(8,4)\).

We have \(n^2+5\equiv4+5=9\pmod{n+2}\). Hence \(n+2\mid9\). Since \(n\ge1\), \(n+2\ge3\), so \(n+2=3\) or \(9\). Thus \(n=1\) or \(n=7\). Both work.

A square modulo \(4\) is \(0\) or \(1\). A sum of two squares modulo \(4\) can only be \(0,1,2\). Residue \(3\) is impossible. Hence \(4k+3\) is not a sum of two squares.

Since \(9\) is invertible modulo \(7\), \(7\mid R_n\) exactly when \(10^n\equiv1\pmod7\). The order of \(10\equiv3\) modulo \(7\) is \(6\). Therefore precisely the multiples of \(6\) work.

We need \(n\equiv1\pmod2\), \(n\equiv1\pmod3\), \(n\equiv2\pmod5\). The first two give \(n\equiv1\pmod6\). The numbers are \(1,7,13,19,\ldots\); residue \(2\) modulo \(5\) gives \(7\). Thus \(n=7\) works.

If \(m\ge n\), then \(2^m-1=(2^{m-n})(2^n-1)+(2^{m-n}-1)\). Hence the GCD is unchanged when \((m,n)\) is replaced by \((m-n,n)\). Repeating the Euclidean algorithm on the exponents leads to \((d,d)\), where \(d=\gcd(m,n)\). The GCD is \(2^d-1\).

Let \(a=x-y\), \(b=x+y\). Then \(ab=2025\), \(a

Assume a solution exists and choose one with minimal \(x+y\). Let \(x\ge y\). View the equation as quadratic in \(x\): \(x^2-3yx+y^2=0\). The second root \(x'=3y-x\) is integral. Since \(x<3y\), it is positive. Also \(x(3y-x)=y^2\); for \(y\le x\le2y\), the left side is at least \(2y^2\), impossible, so \(x>2y\). Hence \(0

If \(2n+1\mid n^2+n+3\), then \(2n+1\mid4(n^2+n+3)\). But \(4(n^2+n+3)=(2n+1)^2+11\). Hence \(2n+1\mid11\). For \(n\ge1\), \(2n+1\ge3\), so \(2n+1=11\), \(n=5\). Check: \(11\mid33\).

Take \(N=(k+1)!\). Then \(N+2,N+3,\ldots,N+k+1\) are \(k\) consecutive numbers. For each \(i=2,3,\ldots,k+1\), the number \(N+i\) is divisible by \(i\), because \(N\) is divisible by \(i\). Also \(N+i>i\), so \(i\) is a nontrivial divisor. Therefore all these numbers are composite.