MHB What Are the Non-Negative Integer Solutions to (xy-7)² = x² + y²?

[anemone]

Here is this week's POTW:

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Find all non-negative integers $x,\,y$ satisfying $(xy-7)^2=x^2+y^2$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Ackbach
2. kaliprasad
3. castor28
4. MegaMoh
5. Olinguito
6. Opalg

Solution from castor28

Spoiler

By symmetry, we may assume that $x\le y$.

If $x=0$, we have $y=7$. Otherwise, we have a proper Pythagorean triple (and $x<y$).

The smallest Pythagorean triples are $(3,4,5)$ and $(5,12,13)$. $(3,4)$ is a solution (since $3\times4-7=5$). We claim that there are no other solutions.

If $5\le x < y$, we have $7<2y$ and $xy-7>3y$. This implies that $(xy-7)^2>9y^2$. On the other hand, since $x<y$, $x^2+y^2<2y^2$, which gives a contradiction and proves the claim.

To summarize, the only solutions are $(0,7)$, $(7,0)$, $(3,4)$ and $(4,3)$.

Alternate solution from Olinguito:

Spoiler

$$(xy-7)^2\ =\ x^2+y^2$$
$\implies\ (x^2-1)(y^2-1)\ =\ 7(2xy-7)+1\ \equiv\ 1\pmod7$

$\implies\ (x^2-1,y^2-1)\ \equiv\ \begin{cases}(1,1)\pmod7 \\ (2,4)\pmod7 \\ (3,5)\pmod7 \\ (4,2)\pmod7 \\ (5,3)\pmod7 \\ (6,6)\pmod7\end{cases}$.

As $n^2-1\not\equiv2,4,5\pmod7$ for any integer $n$, we simply have
$$x^2-1,y^2-1\ \equiv\ 1,6\pmod7$$
$\implies\ x,y\ \equiv\ 0,3,4\pmod7$.

The possibilities reduce to
$$\boxed{(x,y)\ =\ (0,7),(3,4),(4,3),(7,0)}$$
for it is clear that $x,y>7\ \implies\ (xy-7)^2>x^2+y^2$.

Another solution from Opalg:

Spoiler

The equation $(xy-7)^2=x^2+y^2$ is symmetric in $x$ and $y$, so it will be sufficient to find solutions with $0\leqslant x\leqslant y$.

If $x=0$ then the equation becomes $7^2 = y^2$, with the solution $y=7$.

If $x=1$ then it becomes $(y-7)^2 = 1 + y^2$, which simplifies to $7y = 24$, with no integer solution.

If $x=2$ then it becomes $(2y-7)^2 = 4 + y^2$, which simplifies to $3y^2 - 28y + 45 = 0$, again with no integer solutions.

If $x=3$ then it becomes $(3y-7)^2 = 9 + y^2$, which simplifies to $4y^2 - 21y + 20 = (y-4)(4y-5) = 0$, with the solution $y=4$.

Now suppose that $4\leqslant x\leqslant y$. (Notice in passing that this implies $xy\geqslant16$, which means that $xy-7$ is positive.) Then $$(xy-7)^2=x^2+y^2 < x^2 + 2xy + y^2 = (x+y)^2.$$ Take the positive square root of both sides to get $xy-7 < x+y$, from which $(x-1)(y-1) < 8$.

But if $4\leqslant x\leqslant y$ then $(x-1)(y-1) \geqslant (4-1)(4-1) = 9$. That contradicts the previous inequality, showing that that are no solutions with $4\leqslant x\leqslant y$.

In conclusion, there are four solutions of $(xy-7)^2=x^2+y^2$, namely $(x,y) = (0,7),\ (3,4),\ (4,3),\ (7,0).$