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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.

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    $\begingroup$ Please include some of your own thoughts on the problem. Have you seen any similar problems to this one in the past? What sort of tools do you know of that you have available to use here and what was the result of your attempt at using them? If you show that you have at least started the problem and put in effort beyond copy-pasting then you are more likely to receive help. $\endgroup$ – JMoravitz Aug 5 '18 at 23:17
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    $\begingroup$ Well, 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. $\endgroup$ – Dean Yang Aug 5 '18 at 23:24
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    $\begingroup$ The Wikipedia article on Fermat's right triangle theorem says that a square number cannot be congruent, so $1$ is not a congruent number. $\endgroup$ – Ross Millikan Aug 5 '18 at 23:58
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    $\begingroup$ Have you tried a parity argument? What if exactly one of p,q is odd? What if both are odd? $\endgroup$ – Somos Aug 6 '18 at 0:08
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    $\begingroup$ This gives two primitive Pythagorean triples with a common leg, which cannot be. $\endgroup$ – user580373 Aug 6 '18 at 0:58
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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.

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