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Find all tuple $(x,y)$ such that $x,y$ are integers and $(x^2-y^2)^2=20y+1$.

First i see that $x^2-y^2$ is odd and from the fact that a difference between square of two odd is multiple of $8$ and thus $y$ is a multiple of $2$.

Moreover, we have $(x^2-y^2+1)(x^2-y^2-1)=20y$.

Somebody can give some hint! Whether the first information is useful?

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    $\begingroup$ Where does $z$ enter into this? $\endgroup$
    – saulspatz
    Commented Jun 20, 2021 at 16:27
  • $\begingroup$ ops! only $x,y$. Sorry. I will edit now! Thanks! $\endgroup$
    – Dat Tran
    Commented Jun 20, 2021 at 16:31
  • $\begingroup$ The first thing that occurs to me is that $(x^2-y^2)^2\equiv1\pmod{20}$ gives $x^2-y^2\equiv1,9,11,\text{ or }19\pmod{20}$ That gives $4$ congruences to analyze. There are only $6$ squares mod $20$ so it should be easy. I don't know whether it helps, though. $\endgroup$
    – saulspatz
    Commented Jun 20, 2021 at 16:50
  • $\begingroup$ The congruence $x^2-y^2\equiv 1\pmod{20}$ is equivalent to the pair of congruences $x^2-y^2\equiv 1\pmod5$ and $x^2-y^2\equiv 1\pmod4$ by the Chinese remainder theorem; each of these is much easier to analyze than the congruence modulo $20$. $\endgroup$
    – Greg Martin
    Commented Jun 20, 2021 at 16:55

1 Answer 1

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Since $x^2-y^2$ is odd then $x\not=y$ and $$20y+1=(x^2-y^2)^2\geq ((y-1)^2-y^2)^2=(-2y+1)^2=1-4y+4y^2.$$ Hence $$0\geq 4y^2-24y\Leftrightarrow y(y-6)\leq 0 $$ which implies that $0\leq y\leq 6$.

Moreover $20y+1$ is a perfect square, and therefore $y\in\{0,4,6\}$:

  1. if $y=0$ then $(x^2-0)^2=1^2$ and $x=\pm 1$;

  2. if $y=4$ then $(x^2-16)^2=9^2$ and $x=\pm 5$;

  3. if $y=6$ then $(x^2-36)^2=11^2$ and $x=\pm 5$.

Therefore the integer solutions are $(\pm1,0),(\pm5,4),(\pm 5,6).$

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  • $\begingroup$ can you explane why $(x^2-y^2)^2\ge ((y-1)^2-y^2)^2$? I see that it equavalent to $[x^2-(y-1)^2][x^2-y^2-2y+1]\ge 0$. I may not occur! $\endgroup$
    – Dat Tran
    Commented Jun 20, 2021 at 17:21
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    $\begingroup$ The absolute value of the difference of two distinct perfect squares is minimal when the two perfect squares are the squares of two consecutive integers. Therefore $|x^2-y^2|\ge |(y-1)^2-y^2|$. $\endgroup$
    – Robert Z
    Commented Jun 20, 2021 at 17:24
  • $\begingroup$ oh! nice solution! i forgot this property! thank you! $\endgroup$
    – Dat Tran
    Commented Jun 20, 2021 at 17:30
  • $\begingroup$ Are you watching the Italy vs Wales match? :) $\endgroup$
    – Dat Tran
    Commented Jun 20, 2021 at 17:31
  • $\begingroup$ No, but I know the score so far... $\endgroup$
    – Robert Z
    Commented Jun 20, 2021 at 17:35

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