#7 Diophantine Equations I: Factorisation and Bounds

We have \(xy+2x+3y+6=60\), hence \((x+3)(y+2)=60\). Since \(x,y>0\), we need \(x+3\ge 4\), \(y+2\ge 3\). The suitable divisor pairs of \(60\) are \((4,15),(5,12),(6,10),(10,6),(12,5),(15,4),(20,3)\). Thus \((x,y)=(1,13),(2,10),(3,8),(7,4),(9,3),(12,2),(17,1)\). Each pair checks in the original equation.

From \(10x+10y=xy\), we get \(xy-10x-10y=0\), hence \((x-10)(y-10)=100\). Thus \(x=10+d\), \(y=10+\frac{100}{d}\), where \(d\) runs over the positive divisors of \(100\). The answer is \((11,110),(12,60),(14,35),(15,30),(20,20),(30,15),(35,14),(60,12),(110,11)\).

We write \((x-y)(x+y)=105\). The two factors must have the same parity; since the product is odd, both are odd. The factor pairs are \((1,105),(3,35),(5,21),(7,15)\). They give \((x,y)=(53,52),(19,16),(13,8),(11,4)\).

Modulo \(n+5\), we have \(n^2+7\equiv 25+7=32\). Therefore \(n+5\mid 32\). Since \(n+5>5\), we get \(n+5=8,16,32\). Hence \(n=3,11,27\). Direct checking shows that all three values work.

If \(q=2\), then \(p^2=124\), impossible. Thus \(p,q\) are odd, and \(p-q\), \(p+q\) are positive even factors of \(120\). Check \((2,60),(4,30),(6,20),(10,12)\). These give \((p,q)=(31,29),(17,13),(13,7),(11,1)\). The last one is invalid because \(1\) is not prime. The answer is \((31,29),(17,13),(13,7)\).

The equation is equivalent to \(xy-4x-7y=9\). Add \(28\): \((x-7)(y-4)=37\). Since \(37\) is prime, the only factor pairs are \((1,37)\) and \((37,1)\). Thus \((x,y)=(8,41)\) and \((44,5)\).

We get \(8x+8y+8=xy\), so \(xy-8x-8y=8\). Therefore \((x-8)(y-8)=72\). For every positive divisor \(d\mid 72\), \(x=8+d\), \(y=8+\frac{72}{d}\). The solutions are \((9,80),(10,44),(11,32),(12,26),(14,20),(16,17),(17,16),(20,14),(26,12),(32,11),(44,10),(80,9)\).

Rewrite the equation as \((x+y)(x-y)-2(x+y)=21\), hence \((x+y)(x-y-2)=21\). Let \(A=x+y\), \(B=x-y-2\). The positive factor pairs \((A,B)\) are \((1,21),(3,7),(7,3),(21,1)\). The first two give nonpositive \(y\). From \((7,3)\) we get \(x=6,y=1\), and from \((21,1)\) we get \(x=12,y=9\). The answer is \((6,1),(12,9)\).

We have \((x+2)^2-y^2=15\), so \((x+2-y)(x+2+y)=15\). The positive factor pairs with the same parity are \((1,15)\) and \((3,5)\), both odd. They give \((x,y)=(6,7)\) and \((2,1)\). Both check in the original equation.

If \(2n+1\mid n^2+n+10\), then \(2n+1\mid 4(n^2+n+10)\). But \(4n^2+4n+40=(2n+1)^2+39\). Hence \(2n+1\mid 39\). Since \(2n+1>1\) and is odd, \(2n+1=3,13,39\). Thus \(n=1,6,19\), and all three work.

An integer square modulo \(4\) is \(0\) or \(1\). The right side satisfies \(4xy+3\equiv 3\pmod 4\). The left side can be \(0,1,\) or \(2\) modulo \(4\), but never \(3\). Contradiction. Therefore there are no integer solutions.

The equation gives \(y(x+1)=x^2-x-11\). But \(x^2-x-11=(x+1)(x-2)-9\). Hence \(x+1\mid 9\). Since \(x>0\), \(x+1=3\) or \(9\). For \(x=2\), we get \(y=-3\), invalid. For \(x=8\), we get \(9y=45\), so \(y=5\). The answer is \((8,5)\).

Since \(x\le y\le z\), \(1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le \frac{3}{x}\), so \(x\le 3\). If \(x=1\), the remaining sum is \(0\), impossible. If \(x=2\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\), so \((y-2)(z-2)=4\). With \(y\le z\), this gives \((y,z)=(3,6),(4,4)\). If \(x=3\), then \(\frac{1}{y}+\frac{1}{z}=\frac{2}{3}\). Since \(y\ge 3\), the maximum possible left side is \(\frac{2}{3}\), so \(y=z=3\). The answer is \((2,3,6),(2,4,4),(3,3,3)\).

We have \(x^2+y^2=(60-x-y)^2\). Expanding and simplifying gives \(xy-60x-60y+1800=0\), so \((x-60)(y-60)=1800\). Since \(x,y<60\), put \(a=60-x\), \(b=60-y\). Then \(ab=1800\), and the hypotenuse is \(a+b-60\), so we need \(a<60\), \(b<60\), \(a+b>60\). The suitable divisor pairs are \((36,50)\) and \((40,45)\), with swaps. We get the side triples \(24,10,26\) and \(20,15,25\). The answer is \((10,24,26)\) and \((15,20,25)\).

The equation is equivalent to \((x-y)^2=x+y\). Let \(d=|x-y|\). Then \(x+y=d^2\). If \(x\ge y\), then \(x-y=d\), hence \(x=\frac{d^2+d}{2}\), \(y=\frac{d^2-d}{2}\). Positivity requires \(d\ge 2\). Conversely, every \(d\ge 2\) gives positive integers satisfying the equation. Including the swapped order, the answer is \(\left(\frac{d(d+1)}{2},\frac{d(d-1)}{2}\right)\) and its swap, where \(d=2,3,4,\ldots\).

If \(x,y\ge 2\), then \(x^2+y^2-xy-1=(x-y)^2+xy-1\ge 3>0\), so \(x^2+y^2>xy+1\). The divisibility is then impossible because the positive divisor is larger than the positive number. If \(x=1\), the condition becomes \(1+y^2\mid y+1\), which is possible only for \(y=1\). The case \(y=1\) is symmetric. The pair \((1,1)\) indeed works: \(2\mid 2\). The answer is \((1,1)\).

If \(N+1\) is composite, take the block \(N+2,N+3,\ldots,2N+1\). Every prime \(p\le N\) divides the product of any \(N\) consecutive integers, because one of them is divisible by \(p\). Every prime \(p\) with \(N+2\le p\le 2N+1\) appears in the block as one of the factors. The skipped number \(N+1\) is not prime.

If \(N+1\) is prime, take the block \(N+3,N+4,\ldots,2N+2\). Primes \(p\le N\) are again covered by a block of length \(N\). The prime \(N+1\) divides the last term \(2N+2=2(N+1)\). The number \(N+2\) is not prime: if \(N+1\) is prime and greater than \(3\), then \(N+1\) is odd, so \(N+2\) is even and greater than \(2\); the case \(N=2\) is excluded. Primes between \(N+3\) and \(2N+1\) appear in the block. Thus the required block exists in both cases.

From \(x\le y\le z\), we get \(\frac{3}{x}\ge \frac{1}{2}\), so \(x\le 6\). If \(x=1\), the sum is already greater than \(\frac{1}{2}\); if \(x=2\), the remaining fractions would have to sum to \(0\), impossible.

If \(x=3\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{6}\), so \((y-6)(z-6)=36\). With \(y\le z\), this gives \((y,z)=(7,42),(8,24),(9,18),(10,15),(12,12)\).

If \(x=4\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{4}\), so \((y-4)(z-4)=16\). We get \((5,20),(6,12),(8,8)\).

If \(x=5\), then \(\frac{1}{y}+\frac{1}{z}=\frac{3}{10}\). Multiplying and completing a product gives \((3y-10)(3z-10)=100\). With \(y\ge 5\) and \(y\le z\), the only suitable choice is \(3y-10=5\), hence \((y,z)=(5,10)\).

If \(x=6\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\). Since \(y,z\ge 6\), the maximum is \(\frac{1}{3}\), so \(y=z=6\). The answer is \((3,7,42),(3,8,24),(3,9,18),(3,10,15),(3,12,12),(4,5,20),(4,6,12),(4,8,8),(5,5,10),(6,6,6)\).

Let \(D=n^2+3n+3\). From \(n^2\equiv -3n-3\pmod D\), we get \(n^3\equiv -3n^2-3n\equiv 9n+9-3n=6n+9\pmod D\). Thus \(n^3-1\equiv 6n+8\pmod D\), so \(D\mid 6n+8\). For \(n\ge 5\), \(D=n^2+3n+3>6n+8\), impossible. Checking \(n=1,2,3,4\), only \(n=1\) works, since \(7\mid 0\). The answer is \(n=1\).

From \(x\le y\le z\), we have \(\frac{3}{x}\ge \frac{2}{3}\), hence \(x\le 4\). The value \(x=1\) is impossible because \(\frac{1}{x}=1>\frac{2}{3}\).

For \(x=2\): \(\frac{1}{y}+\frac{1}{z}=\frac{1}{6}\), so \((y-6)(z-6)=36\). With \(y\le z\), this gives \((y,z)=(7,42),(8,24),(9,18),(10,15),(12,12)\).

For \(x=3\): \(\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\), so \((y-3)(z-3)=9\). We get \((y,z)=(4,12),(6,6)\).

For \(x=4\): \(\frac{1}{y}+\frac{1}{z}=\frac{5}{12}\). Multiplying and completing a product gives \((5y-12)(5z-12)=144\). We need \(y\ge 4\), \(y\le z\), hence \(5y-12\le 12\), and this divisor must be congruent to \(3\) modulo \(5\). Among divisors of \(144\) not exceeding \(12\), the possibilities are \(3\) and \(8\); \(3\) gives \(y=3\), invalid, while \(8\) gives \(y=4,z=6\). The answer is \((2,7,42),(2,8,24),(2,9,18),(2,10,15),(2,12,12),(3,4,12),(3,6,6),(4,4,6)\).