Mathematica competitions in Croatia

Let us rewrite the equation: $$x^4+68=4y^4$$ $$x^4-4y^4=-68$$ $$(x^2-2y^2)(x^2+2y^2)=-68$$

Now, $-68$ factors as $(-1)\times 68$, $(-2)\times 34$, $(-4)\times 17$, $(-17)\times 4$, $(-34)\times 2$, $(-68)\times 1$ and their negatives. We consider all pairs $(a,b)$ such that $a\times b=-68$.

Let $x^2-2y^2=a$, $x^2+2y^2=b$, so $a+b=2x^2$, $b-a=4y^2$.

Thus, $$x^2=\frac{a+b}{2}$$ $$y^2=\frac{b-a}{4}$$

We need $x^2$ and $y^2$ to be integers, so $a+b$ is even and $b-a$ is divisible by $4$.

Let us check all possible pairs $(a,b)$:

1. $a=-1$, $b=68$ - $a+b=67$ (odd), $b-a=69$ (not divisible by $4$) 2. $a=-2$, $b=34$ - $a+b=32$ (even), $b-a=36$ (divisible by $4$) - $x^2=16$, $y^2=9$ - $x=\pm 4$, $y=\pm 3$ 3. $a=-4$, $b=17$ - $a+b=13$ (odd), $b-a=21$ (not divisible by $4$) 4. $a=-17$, $b=4$ - $a+b=-13$ (odd), $b-a=21$ (not divisible by $4$) 5. $a=-34$, $b=2$ - $a+b=-32$ (even), $b-a=36$ (divisible by $4$) - $x^2=-16$ (not a square), $y^2=9$ 6. $a=-68$, $b=1$ - $a+b=-67$ (odd), $b-a=69$ (not divisible by $4$)

Now, the negatives: 7. $a=1$, $b=-68$ - $a+b=-67$ (odd), $b-a=-69$ (not divisible by $4$) 8. $a=2$, $b=-34$ - $a+b=-32$ (even), $b-a=-36$ (divisible by $4$) - $x^2=-16$ (not a square), $y^2=-9$ (not a square) 9. $a=4$, $b=-17$ - $a+b=-13$ (odd), $b-a=-21$ (not divisible by $4$) 10. $a=17$, $b=-4$ - $a+b=13$ (odd), $b-a=-21$ (not divisible by $4$) 11. $a=34$, $b=-2$ - $a+b=32$ (even), $b-a=-36$ (divisible by $4$) - $x^2=16$, $y^2=-9$ (not a square) 12. $a=68$, $b=-1$ - $a+b=67$ (odd), $b-a=-69$ (not divisible by $4$)

So, the only valid case is $a=-2$, $b=34$: $$x^2=16⟹x=\pm 4$$ $$y^2=9⟹y=\pm 3$$

Check in the original equation: $$x^4+68=4y^4$$ $$256+68=4\times 81$$ $$324=324$$

Thus, all integer solutions are: $$(x,y)=(4,3),(4,-3),(-4,3),(-4,-3)$$