- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,894

-----

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

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter anemone
- Start date

- Status
- Not open for further replies.

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,894

-----

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

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
- Admin
- #2

- Feb 14, 2012

- 3,894

1. Ackbach

2. kaliprasad

3. castor28

4. MegaMoh

5. Olinguito

6. Opalg

Solution from castor28

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 :

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

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

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 :

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 $(x,y) = (0,7),\ (3,4),\ (4,3),\ (7,0).$

- Status
- Not open for further replies.