Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $

Question

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.

Answer 1

HINT: From the second equation we get that $y \mid 30$. As $30$ has $16$ divisors (both negative and positive) you can take a look at each of them separately. Also this uniquely determines $x$, so you just have to plug it in the first equation to check if the pair is a solution.

Answer 2

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.

Answer 3

HINT.-If you want to have another way then you can do as follows: $$y^2+xy-30=0\iff 2y=-x\pm\sqrt{x^2+120}$$ the only possibilities are $x=\pm1\text{ or } \pm7$ and $x=\pm1$ is not compatible. This proves that $(7,3)$ and $(-7,-3)$ are the only solutions.

Answer 4

Continuing your calculation,

$(\dfrac{30-y^2}y)^2=y^4+y^2-41,y≠0 $

$⇒y^4-60y^2+900=y^2(y^4+y^2-41)$

$⇒y^6+19y^2-900=0$

$⇒y^2(y^4+19)=900$

This give integer solution $y^2=9$.

Answer 5

Here is yet another way in. Note that $y$ must be a factor of $30$ as must $x+y$ (both from the second equation).

You can rewrite the first as $(y^2+x)(y^2-x)=xy+11=$ and use the second $=41-y^2$

Now note also that changing the signs of both $x$ and $y$ still gives a solution, so we can assume $y\gt 0$.

If both $x$ and $y$ are positive $xy+11\gt 0$ and $y\le 6$

If $y$ is positive and $x$ is negative, then $x+y\gt 0$ from the second equation so $y\gt -x$, and since $y^2 \ge y$ we have also that $y^2\gt -x$, whence $(y^2+x)(y^2-x)\gt 0$ and $y\le 6$

So positive $y$ is restricted to $y\le 6$ and the possibilities are easily tested.