-1

Let $x,y$ be nonzero integers.

I could not find primes apart from $5$ of the form $x^4 + 4 y^4$.

Why is that ?

I know that if x and y are both not multiples of $5$ then it follows from fermat's little.

And if both are multiples of $5$ the expression is trivially a multiple of $5$.

But beyond that im stuck.

Biquadratic reciprocity ??

Im not sure.

mick
  • 15,946
  • 5
    $x^4 + 4y^4 = (x^4 + 4x^2y^2 + 4y^4) - 4x^2y^2 = (x^2 + 2y^2)^2 - (2xy)^2$ – Daniel Fischer Aug 14 '15 at 11:29
  • 1
    Using Sophie Germain Identity, $(x^4+4y^4)=(x^2+2y^2)^2-(2xy)^2 = (x^2+2xy+2y^2)\cdot (x^2-2xy+2y^2) $ – juantheron Aug 14 '15 at 11:30
  • 2
    The technique (Sophie Germain / Aurifeuillian factorization) has appeared on our site so many times that it is pointless to pick one. Start here (note that even that is closed as a dup), and examine the list of Related questions. True, many instances consider a special case. – Jyrki Lahtonen Aug 14 '15 at 11:49
  • 2
    Also show $x^2-2xy+2y^2 = (x-y)^2+y^2 > 1$ unless $x=y=\pm1$. – Marconius Aug 14 '15 at 12:46

2 Answers2

1

Hint

$$x^4+4y^4=(x^2+2y^2)^2-(2xy)^2=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$$

math110
  • 93,304
1

use that $$(x^2)^2+(2y^2)^2=(x^2)^2+(2y^2)^2+4x^2y^2-4x^2y^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$$

Dr. Sonnhard Graubner
  • 95,283