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I need solution check to see if I overlooked something:

If $x^4+y^4=z^2$ has an integer solution then $(\frac{x}{y})^4+1=(\frac{z}{y^2})^2$ has a solution in rationals.

Second equation is equivalent to $X^4+1=Y^2$ which can be written as cubic Weierstrass form $$v^2=u^3-4u$$ where $u=\frac{2(Y+1)}{X^2}$ and $v=\frac{4(Y+1)}{X^3}$.

By using SAGE we can see that Mordell-Weil group of this elliptic curve is empty (SAGE returns empty set, but I’m not sure if that is enough to conclude this), so we can conclude that there are no rational solutions to the second equation and no integer solution to the first one.

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**edit:** No integer solution in **positive integers**.

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SAGE 6.7 implements an algorithm due to Peth, et al. (See A. Peth, H. G. Zimmer, J. Gebel, E. Herrmann, Computing all S-integral points on elliptic curves, Math. Proc. Camb. Philos. Soc. 127 (1999), No.3, 383-402) which is considered robust. I’ve had papers accepted which invoked SAGE, and it was trusted with finding all the integer points on the elliptic curves. So the answer to your question is “Yes”. But of course, some reader may have doubts.

There are no positive integer solutions to $x^4+y^4=z^2.$ By contradiction suppose $(x,y,z)$ is a positive solution with the least possible $z.$

If a prime $p$ divides $x$ and $y$ then $p^4$ divides $z^2,$ so $p^2$ divides $z.$ But then $(x’,y’,z’)=(x/p,y/p,z/p^2)$ is a positive solution with $z'<z,$ contradicting the minimality of $z.$

Therefore $x,y$ are co-prime. So $(x^2,y^2,z)$ is a Fundamental Pythagorean triplet. So there exist co-prime positive integers $m,n$, not both odd, with $\{x^2,y^2\}=\{m^2-n^2, 2mn\}$ and $z=m^2+n^2.$ WLOG $x^2=m^2-n^2$ and $y^2=2mn.$ Note that $x$ is odd and $y$ is even.

Now $x^2+n^2=m^2.$ A prime divisor of both $x$ and $n$ must divide $m,$ but $\gcd(m,n)=1.$ So $x,n$ are co-prime and $(x,n,m)$ is a Fundamental Pythagorean triplet. So there are co-prime positive integers $a,b$ with $\{a^2-b^2, 2ab\}=\{x,n\}$ and $a^2+b^2=m.$ Since $x$ is odd we have $x=a^2-b^2$ and $n=2ab.$

Now $y^2=2mn=2(a^2+b^2)(2ab)=4(a^2+b^2)ab.$ But $\gcd (a,b)=1$ so the members of $\{a,b, a^2+b^2\}$ are pair-wise co-prime and their product $(y/2)^2$ is a square. Therefore each of $a,b,a^2+b^2$ is a square: There are positive integers $d,e,f$ with $a=d^2,b=e^2,a^2+b^2=f^2.$ Therefore $$ f^2=a^2+b^2=d^4+e^4.$$ But $f^2=a^2+b^2=m<m^2+n^2=z,$ so $f\leq f^2<z.$ This contradicts the minimality of $z.$

The following is *not an answer*, almost like **a lengthy version of a comment**.

Say $a=x^2$ and $b=y^2$.

Then the equation becomes $a^2+b^2=z^2$.

Now we can conclude that $a,b,z$ will of the following forms courtesy this proof:

$$a=d\cdot 2pq$$

$$b=d\cdot (p^2-q^2)$$

$$z=d\cdot (p^2+q^2)$$

where $d=\mathrm{gcd}(a,b,z)$ and $p,q \in \mathbb{N}$.

Now $a$ and $b$ are $2$ perfect squares.

So the problem now reduces to this:

Prove that numbers having the form $d\cdot 2pq$ and $d\cdot (p^2-q^2)$ cannot be perfect squares.

See if this helps.

**P.S.** If you still want some alternate method to find solutions of $a^2+b^2=z^2$, you can look up this link.

I think you equation has infinitely many integer solutions.

For example $\{(a,0,a^2)|a\in\mathbb Z\}$.

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