Given that $7x^2 - 40xy + 7y^2 = (|(x - y)| + 2)^3$ And $x-y\equiv a(\mod 13)$ Solve for $a$

Question

Given that $$7x^2 - 40xy + 7y^2 = (|(x - y)| + 2)^3$$ and $x-y\equiv a(\mod 13)$. Solve for $a$

---

Im not sure how to approach this, hints and solutions would be appreciated.

Answer 1

Well everything $\mod 13$:

$|x-y|^2 = x^2 - 2xy+y^2$

$7|x-y|^2 = 7x^2 -14xy + 7y^2 \equiv 7x^2-40xy + 7y^2$ so

$7|x-y|^2\equiv (|x-y|+2)^3$

So we must solve $7a^2 \equiv (a + 2)^3\pmod{13}$

....

Now $(\pm k)^2 \equiv 0,1,4,-4,3,-1,-3$ for $k=0,1,2,3,4,5,6$

And $7(\pm k)^2 \equiv 0,-6,2,-2,-5,6,2$

While $k^3\equiv 0,1,-5,1,-1,-5,-5,5,5,1,-1,5,-1$

So we must have $a\equiv \pm 4$ and $a+2\equiv 2,5,6$.

So $a\equiv 4$