# Contest math integer doublet equation

Can anyone help me with this?

Find all ordered pairs $(x, y)$ of positive integers $x$, $y$ such that $$x^2 + 4y^2 = (2xy − 7)^2$$

## 2 Answers

We have $$\left(2y(x-1)-7\right)\left(2y(x+1)-7\right)=x^2.$$ Obviously $x \ne 1$ and $x$ is odd, i.e. $x \ge 3$. Since $y \ge 3$ implies $2y(x-1)-7 \ge x+5x-13 > x$, you only need to check $y=1$ and $y=2$.

• How does y greater or equal to 3 imply that? – Yadnarav3 Sep 12 '13 at 22:10
• And how did you get it in that first form? Or rather, what led you to it? – Yadnarav3 Sep 12 '13 at 22:12
• @Yadnarav3 This comes directly from factoring (difference of squares) on $(2xy - 7)^2 - 4y^2 = x^2$. – 6005 Sep 13 '13 at 4:14
• @Yadnarav3 Whenever you have lots of squares like this, difference of squares is a natural and potentially very useful way to proceed. – 6005 Sep 13 '13 at 4:15
• What led you to y greater or equal to 3 – Yadnarav3 Sep 13 '13 at 18:00

$$x^2 + 4y^2 - 4xy +4xy = (2xy-7)^2$$ $$(x+2y)^2 = (2xy-7)^2 + 4xy$$ $$= 4x^2y^2 - 28xy +4xy +49$$ $$= 4x^2y^2 - 24xy +49$$ $$= (4x^2y^2 - 24xy +36) +13$$ $$(x+2y)^2 - (2xy -6)^2 +13$$ $$(x+2y +2xy -6) (x+2y - 2xy +6) = 13$$ The product of these two terms can only be 13 and 1 either $(x+2y +2xy -6) = 13$ and $(x+2y - 2xy +6) = 1$ or $(x+2y+ 2xy - 6) = 1$ and $(x+2y - 2xy + 6) = 13$

Case i) $x+2y+2xy - 6 = 13$ $x+2y - 2xy + 6) = 1$ Adding these two We get $2x+4y = 14$ Subtracting We get $4xy -12 = 12$ This gives $xy = 6$

Case ii) $x+2y+2xy - 6 = 1$ $x+2y - 2xy + 6 = 13$ Adding these two We get $2x+4y = 14$ Subtracting We get $4xy -12 = -12$ This gives $xy = 0$

Case ii) is not possible because x and y postive integers

Case i) will lead to possible combinations of $x= 6$ or $y =1$, or $x=3$ and $y = 2$ or $x=2$ and $y = 3$ or $x= 1$ and $y = 6$ as xy are positive integers

Substituting back you will see that the ordered pair(x,y) is (3,2) is the only solution

Thanks

Satish