Given that $p$ and $q$ have no common factors, how can I prove that $p^4 + 4q^4 \not =z^2$, if $p,q$ and $z$ have to be positive integers?
From the comments:
I'm working on the congruent number problem and I'm trying to prove 1 isn't a congruent number and have reduced the problem to this. With regards, to what I've seen before- I've seen a proof of thermat's last theorem for the case $n=4$, where they proved that $p^4 + q^4$ is never a perfect square.
It's an explanation of the Ross Millikan's comment.
If $\gcd(p,q)=1$ and $p^4+4q^4=z^2$ then there are $m$ and $n$ with different parity,
for which $m>n$, $\gcd(m,n)=1$, $$p^2=m^2-n^2$$ and $$2q^2=2mn.$$ Thus, $m=x^2$, $n=y^2$ and we get that the equation $$p^2=x^4-y^4$$ has solutions in natural numbers.
But $$(x^4-y^4)^2+(2x^2y^2)^2=(x^4+y^4)^2,$$ which says that $$(x^4-y^4,2x^2y^2,x^4+y^4)$$ is a Pythagorean triple and the area of this right-angled triangle it's $$\frac{(x^4-y^4)\cdot2x^2y^2}{2}=(pxy)^2,$$ which is impossible.