Diophantine Equations I: Factorisation and Bounds

Divide by \(2\): \(2x+3y=5\). One solution is \(x=1\), \(y=1\). Hence all solutions are \(x=1+3t\), \(y=1-2t\), where \(t\in\mathbb Z\). Check: \(2(1+3t)+3(1-2t)=5\).

From \(2x=18-3y\), the number \(y\) must be even. Since \(y>0\) and \(3y<18\), we have \(y=2\) or \(y=4\). Then \(x=6\) or \(x=3\). The solutions are \((6,2)\), \((3,4)\).

The positive divisors of \(24\) are \(1,2,3,4,6,8,12,24\). For each \(x\), we have \(y=\frac{24}{x}\). Thus the pairs are \((1,24)\), \((2,12)\), \((3,8)\), \((4,6)\), \((6,4)\), \((8,3)\), \((12,2)\), \((24,1)\).

We have \((x-y)(x+y)=15\). The positive factor pairs with the same parity are \((1,15)\) and \((3,5)\). They give \((x,y)=(8,7)\) and \((4,1)\). These are all solutions.

A square modulo \(4\) is \(0\) or \(1\). Therefore a difference of two squares modulo \(4\) can have residue \(0,1\), or \(3\), but not \(2\). The left-hand side cannot equal \(2\). No solutions exist.

One solution is \(x=-3\), \(y=2\), since \(7(-3)+11\cdot2=1\). All solutions are \(x=-3+11t\), \(y=2-7t\), where \(t\in\mathbb Z\).

Modulo \(5\), \(8y\equiv67\), so \(3y\equiv2\pmod5\). Hence \(y\equiv4\pmod5\). Since \(8y<67\), we have \(y\le8\), so \(y=4\). Then \(x=\frac{67-32}{5}=7\). The answer is \((7,4)\).

We get \((x+1)(y+1)=48\). Thus \(x+1\) and \(y+1\) are divisors of \(48\), both at least \(2\). The solutions are \((1,23)\), \((2,15)\), \((3,11)\), \((5,7)\), \((7,5)\), \((11,3)\), \((15,2)\), \((23,1)\).

We have \(xy-2x-5y=0\), hence \((x-5)(y-2)=10\). The positive divisors of \(10\) are \(1,2,5,10\). Therefore the solutions are \((6,12)\), \((7,7)\), \((10,4)\), \((15,3)\).

\((x-y)(x+y)=56\). The two factors must have the same parity, so both are even. The suitable pairs are \((2,28)\) and \((4,14)\). They give \((x,y)=(15,13)\) and \((9,5)\).

From \(8x+8y=xy\), we get \(xy-8x-8y=0\), hence \((x-8)(y-8)=64\). Therefore \(x=8+d\), \(y=8+\frac{64}{d}\), where \(d\) runs over the positive divisors of \(64\).

Modulo \(3\), we get \(x^2\equiv2\pmod3\). But a square modulo \(3\) can have only residue \(0\) or \(1\). This is a contradiction, so no solutions exist.

We have \(xy+2x+3y+6=60\), so \((x+3)(y+2)=60\). Since \(x\ge1\), \(y\ge1\), we need \(x+3\ge4\), \(y+2\ge3\). Checking the divisors of \(60\), we get \((1,13)\), \((2,10)\), \((3,8)\), \((7,4)\), \((9,3)\), \((12,2)\), \((17,1)\).

\((x-y)(x+y)=68\). The only suitable positive factor pair with the same parity is \((2,34)\). The pair \((4,17)\) has different parity. Thus \(x=\frac{2+34}{2}=18\), \(y=\frac{34-2}{2}=16\). The answer is \((18,16)\).

Multiplying gives \(10x+10y=xy\). Then \(xy-10x-10y=0\), and \((x-10)(y-10)=100\). Hence \(x=10+d\), \(y=10+\frac{100}{d}\), where \(d\) is a positive divisor of \(100\). This describes all solutions.

Rewrite as \((x+1)^2-y^2=21\). Then \((x+1-y)(x+1+y)=21\). The positive factor pairs are \((1,21)\) and \((3,7)\). They give respectively \((x,y)=(10,10)\) and \((4,2)\). There are no other solutions.

We have \(xy-x-2y+2=19\), so \((x-2)(y-1)=19\). Since \(19\) is prime, the possible factor pairs are \((1,19)\) and \((19,1)\). Thus \((x,y)=(3,20)\) and \((21,2)\).

A square modulo \(8\) is \(0,1\), or \(4\). A sum of two such residues can be \(0,1,2,4,5\) modulo \(8\), but not \(6\). The right-hand side has residue \(6\), a contradiction.

We get \(xy-x-y+1=2\), so \((x-1)(y-1)=2\). Hence \((x-1,y-1)=(1,2)\) or \((2,1)\). The solutions are \((2,3)\), \((3,2)\).

From \(xy-4x-4y=5\), we get \((x-4)(y-4)=21\). The positive divisors of \(21\) are \(1,3,7,21\). This gives \((5,25)\), \((7,11)\), \((11,7)\), \((25,5)\). Negative factors do not give positive \(x,y\).

The equation is equivalent to \((x-y)^2=x+y\). If \(x=y\), then \(0=2x\), impossible for positive \(x\). Suppose \(x>y\) and let \(d=x-y\). Then \(d^2=2y+d\), so \(y=\frac{d^2-d}{2}\), \(x=\frac{d^2+d}{2}\). We need \(d\ge2\). By symmetry, all solutions are \(\left(\frac{d(d+1)}{2},\frac{d(d-1)}{2}\right)\) and \(\left(\frac{d(d-1)}{2},\frac{d(d+1)}{2}\right)\), where \(d\ge2\) is an integer.

Let \(x\le y\). Then \(xy\le x+y+2\). If \(x\ge3\), then \(xy\ge3y\), while \(x+y+2\le2y+2\), impossible for \(y\ge3\). Thus \(x=1\) or \(x=2\). For \(x=1\), \(y\mid y+3\), so \(y\mid3\), and \(y=1,3\). For \(x=2\), \(2y\mid y+4\), so \(y\le4\); checking \(y=2,3,4\) leaves only \(y=4\). By symmetry, the solutions are \((1,1)\), \((1,3)\), \((3,1)\), \((2,4)\), \((4,2)\).

Factor: \((x-y)(x+y)=2(x+y)+15\). Hence \((x+y)(x-y-2)=15\). Let \(A=x+y\), \(B=x-y-2\). Then \(AB=15\), \(A>0\), \(B>0\). Check the pairs \((15,1)\), \((5,3)\), \((3,5)\), \((1,15)\). Since \(y=\frac{A-B-2}{2}\), a positive \(y\) occurs only for \((A,B)=(15,1)\). Then \(y=6\), \(x=9\). The answer is \((9,6)\).

The equation is symmetric, so assume \(x\ge y\). Then \(x^2\le x^2+xy+y^2=x+y+30\le2x+30\). Thus \(x^2-2x-30\le0\), so \(x\le6\).

Check \(x=1,2,3,4,5,6\), viewing the equation as a quadratic in \(y\). For \(x=1\), \(y^2=30\), no solution. For \(x=2\), \(y^2+y-28=0\), no positive integer \(y\). For \(x=3\), \(y^2+2y-24=0\), so \(y=4\), contradicting \(x\ge y\). For \(x=4\), \(y^2+3y-18=0\), so \(y=3\). For \(x=5\), \(y^2+4y-10=0\), no integer \(y\). For \(x=6\), \(y^2+5y=0\), no positive \(y\). Hence under \(x\ge y\) only \((4,3)\) appears. By symmetry, \((3,4)\) also appears. The solutions are \((4,3)\), \((3,4)\).