I am trying to prove that $X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers. I know there are similar questions on MS, but that minus signs before the $Z$ gives me a hard time.

For the moment, I converted the equation to the equivalent $X^4 + Z^2 = 4Y^2$, and I supposed that $(x,y,z)$ is a primitive solution for this equation. Thus we have $(x^2)^2 +z^2 = (2y^2)^2$.

Now $(2y^2)^2$ is even, it tells us that $x$ and $z$ must have the same parity. But we know by Diophante's equation that there exist $a$ and $b$ with $a > b$, $(a, b) = 1$. If $a \not \equiv b \pmod 2$, we have $x^2 = a^2 - b^2, z = 2ab$ and $2y^2 = a^b + b^2$. But this is impossible, because of $2y^2 = a^2 + b^2$, $a$ and $b$ should have the same parity, which is not our hypothesis.

If $a$ and $b$ have the same parity, they are both odd (otherwise they are not coprime) and we have $x^2 = \frac{a^2 - b^2}{2}$, $z = ab$ and $2y^2 = \frac{a^2 + b^2}{2}$. But once again, as $(a, b) = 1$ and are both odd, they have no factor "2" in common thus $2y^2 = \frac{a^2 + b^2}{2}$ is impossible.

Therefore the equation $X^4 - 4Y^4 = -Z^2$ has no solution in $\mathbb{N} \setminus \{0\}$.

What do you think of this argument? My first idea was to use the infinite descent, but in this case I did not manage to do it. My second idea was to prove that it was equivalent to another equation such as $X^4 + Y^4 = Z^2$ which we know has no solutions such that $XYZ \neq 0$ but I did not have more success.

  • $\begingroup$ Just a thought: did you consider difference of squares on the left side? $\endgroup$ – barrycarter Jan 27 '16 at 2:07
  • $\begingroup$ What do you mean by difference of squares? $\endgroup$ – Laurent Hayez Jan 27 '16 at 13:02
  • $\begingroup$ See @ArnieDris answer. $\endgroup$ – barrycarter Jan 27 '16 at 15:48
  • $\begingroup$ @LaurentHayez you should consider accepting one of the provided answers if they address your concerns, or asking a fallow up question if they don't. $\endgroup$ – Stella Biderman Feb 1 '16 at 23:57

Your argument is correct, and no quicker method is springing to mind to solve this.


For another way of showing your required conclusion, I give you the following:

Hint: $$Z^2 = 4Y^4 - X^4 = (2Y^2 + X^2)(2Y^2 - X^2)$$

Can you take it from here?

  • $\begingroup$ How would that help? $\endgroup$ – individ Jan 27 '16 at 6:28
  • $\begingroup$ Set $A = 2Y^2 + X^2$ and $B = 2Y^2 - X^2$. If $\gcd(A, B) = 1$, then since $Z^2 = AB$, $A$ and $B$ are perfect squares. Then consider the case $\gcd(A, B) > 1$ separately. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 27 '16 at 6:34
  • $\begingroup$ Additionally, $A$ and $B$ are determined uniquely by the quadratic equation $x^2 - {4Y^2}x + Z^2 = 0$, which has discriminant $16{Y^4} - 4{Z^2}$ (and happens to be equal to $4X^4$). $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 27 '16 at 6:39
  • $\begingroup$ No. Come to an equivalent equation. Have to solve a similar equation. $\endgroup$ – individ Jan 27 '16 at 6:46
  • $\begingroup$ If $\gcd(A, B) = 1$ then you fall back on an equation of the form $2y^2 - x^2 = a^2$, this seems possible to prove there are no solutions (although I did not try it), but if $\gcd(A, B) > 1$, what would you do? $\endgroup$ – Laurent Hayez Jan 27 '16 at 13:01

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