A number theory problem by Vaibhav Prasad

$(x-y)^2+2y^2=43$

$\implies x^{2} - 2xy + 3y^2 -43 = 0$

Let's consider this equation as a quadratic polynomial in x.

For roots of this polynomial to be integral, discriminant should be non-negative.

Discriminant of this polynomial = $(-2y)^2 -4(1)(3y^2 -43) \geq 0$

$\implies 5> y > -5$

Now we have bounded the possible values of y ie, possible values for y = $4,3,2,1,0,-1,-2,-3,-4$

Also, by a little modular arithmetic, we can show that y is odd.

Now remaining possible values of y = $3,1,-1,-3$

Out of these possible values, only y = 3 and y = -3 give integral values of x.

y = 3 gives two distinct integral values of x.

Similarly y = -3 also give two distinct integral values of x.

Therefore total number of possible ordered pairs = $\boxed{4}$

No need to expand the LHS, just limit the possibilities by using the properties of squares and the finite number of squares less than 43: $\ 0 < (x-y)^2 \ , \ 2y^2 < 43 \ (no \ equality \ as \ x \ne y )$ and the squares available are 1,4,9 and 16. Now let $\ y^2 = 1,4,9,16$ such that: $\ (x-y)^2 = 43 - 2y^2 = 41 \ , \ 35 \ , \ 25 \ , \ 11$ $\implies\ (x-y)^2 = 25 \ and \ y^2 = 9$ $\therefore\ (x,y) = (-2, 3) \ , \ (8,3) \ , \ (-8, -2) \ , \ (2, -3)$

Let $z=x-y$ :

$(x-y)^2+2y^2=43$

$z^2+2y^2=43$

$z=\pm\sqrt{43-2y^2} \Longrightarrow 43-2y^2 \geq 0$

$y \leq 4 <\sqrt{\frac{43}{2}} \approx 4.6$

Checking values of $y$ between $1$ and $4$ that make $43-2y^2$ a square number:

$43-2(1)^2=41$

$43-2(2)^2=35$

$43-2(3)^2=25=\boxed{5^2}$

$43-2(4^2)=11$

Therefore only $y=\pm3$ yields a square number, giving us the solutions $z=\pm5\Longrightarrow x=\{\pm2, \pm8\}$ . This yields four possible solutions:

$(8,3), (-2,3), (2,-3), (-8,-3)$

Therefore our answer must be $\boxed{4}.$

The only possible RHS can be (5)^2 + 2(3)^2 Hence ordered pairs can be represented by the set {(8,3),(-2,3),(2,-3),(-8,-3)}