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.
Hello,
Consider this equation modulo 4 :
3=-1 mod 4
so we're back to $\displaystyle x^4+y^4=-z^4 \Leftrightarrow x^4+y^4+z^4=0$
the only solution to this is $\displaystyle x=y=z=0 \bmod 4$
so x=4x', y=4y', z=4z', for some integers x', y', z' different from 0
thus the equation is $\displaystyle 256x'^4+256y'^4=3 \cdot 256 z'^4$
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)
Also asked and answered here: http://www.mathhelpforum.com/math-he...rithmetic.html