An Olympiad question about equality

Question

How many $(x, y)$ positive integer pairs have $y^2-x^2=2y+7x+4$ equality?

I can't solve this Olympiad question.

$\textbf{Solution:}$ If above equality is regulated,we obtain $(2x+2y-5)(2x-2y+9)=29$. And, $29$ is prime we can easily find solution.

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• $\textbf{1-)}$ How can we regulate it to obtain $(2x+2y-5)(2x-2y+9)=29$?

• $\textbf{2-)}$ Are there the general method for this kind of questions?

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Thank you for help...

Answer 1

It's just completing the square,

$$y^2-2y-(x^2+7x)=4$$

$$y^2-2y+1-1-\left(x^2+7x + \left(\frac72\right)^2 \right)+\left(\frac72\right)^2=4$$

$$(y-1)^2-1-\left(x+ \frac72 \right)^2+\left(\frac72\right)^2=4$$

Multiply everything by $4$.

$$(2y-2)^2-4-\left(2x+ 7 \right)^2+49=16$$

Answer 2

Let the factors $u:=y+x$ and $v:=y-x$ appear.

$$uv=\frac{9u-5v}2+4.$$

Then,

$$\left(2u+5\right)\left(2v-9\right)=61.$$

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Update to please Misha Lavro:

The factorization of the LHS hints the introduction of these intermediate variables: as the RHS is linear, we know that we can transform to $uv=au+bv+c$, which factors as $(u-p)(u-q)=r$.