1. ## Proof

Deduce from that the equation x^4+y^4 =3z^4 has no solutions in Z (intergers) except the trivial one (x;y; z) = (0;0;0).

I Know it got something to with Fermat’s Last Theorem, but i just can't seem to do this question, can someone please help.

2. Hello,
Originally Posted by nerdo
Deduce from that the equation x^4+y^4 =3z^4 has no solutions in Z (intergers) except the trivial one (x;y; z) = (0;0;0).

I Know it got something to with Fermat’s Last Theorem, but i just can't seem to do this question, can someone please help.
Consider this equation modulo 4 :
3=-1 mod 4

so we're back to $x^4+y^4=-z^4 \Leftrightarrow x^4+y^4+z^4=0$
the only solution to this is $x=y=z=0 \bmod 4$
so x=4x', y=4y', z=4z', for some integers x', y', z' different from 0

thus the equation is $256x'^4+256y'^4=3 \cdot 256 z'^4$

divide by 256 and you'll get the initial equation again.

this is the method of infinite ascent (if I'm not mistaking), similar to Fermat's infinite descent, and this means that there is no other solution than (0,0,0)

3. Originally Posted by nerdo
Deduce from that the equation x^4+y^4 =3z^4 has no solutions in Z (intergers) except the trivial one (x;y; z) = (0;0;0).

I Know it got something to with Fermat’s Last Theorem, but i just can't seem to do this question, can someone please help.