<s>[INST] Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $ 
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $

Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4  +y^2 $
$x^2 = y^4  +y^2 - 41$
$x^2 -49 = y^4  +y^2 - 41 - 49$
$(x+7)(x-7)= (y^2 +10) (y^2-9)$
And from here you get that two pairs can be: $(3,7)$ and $(-3,-7)$.
I think I've made some progress but don't know if those are the only two pairs that satisfy the equation.
Also I would like to see a different way of solving it since I think subtracting that $49$ was just luck.
 [/INST] After you render $x^2=y^4+y^2-41$, complete the square on the right side.  Multiplying by $4$ to eliminate fractions then gives
$(2x)^2-(2y^2+1)^2=-165$
Then from the difference of squares factorization the following are both divisors of $165$,  with product $-165$:
$2x+2y^2+1$
$2x-2y^2-1$
Their sum is then $4x$ which must be the sum of two divisors of $165$, one positive and one negative, with the proper product; thus
$4x=\pm(165-1)=\pm 164, x=\pm 41$
Or
$4x=\pm(55-3)=\pm 52, x=\pm 13$
Or
$4x=\pm(33-5)=\pm 28, x=\pm 7$
Or
$4x=\pm(15-11)=\pm 4, x=\pm 1$
And we try all these to find integer roots for $y$, finding that  only $x=\pm 7$ makes it.
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