Let`s first we try to find such $ a,b,c,d\in N$, that $ a^4 + b^4 + 4c^4 = d^4$. Let $ m,n,p\in N$ and $ a = 2^n$, $ b = 2^m - 1$, $ c = 2^p$, $ d = 2^m + 1$. Then we get: \[ 2^{4n} + 2^{4p + 2} = (2^m + 1)^4 - (2^m - 1)^4 \] \[ 2^{4n} + 2^{4p + 2} = 2^{m + 3}(2^{2m} + 1) \] Let $ p \geq n$, then: \[ 2^{4n}(2^{4p + 2 - 4n} + 1) = 2^{m + 3}(2^{2m} + 1) \] Now we find such $ m = m(n)$ and $ p = p(n)$. Let: $ 4n = m + 3$ and $ 4p + 2 - 4n = 2m$. From first we get: $ m = 4n - 3$. From second: $ p = 3n - 2$. Finally for $ n\in N$: \[ (a,b,c,d) = (2^n, 2^{4n - 3} - 1, 2^{3n - 2}, 2^{4n - 3} + 1) \] Return to our problem: Find $ x,y,z\in Q$, that $ x^4 + y^4 + 4z^4 = 1$. If we put $ x = \frac {a}{d}, y = \frac {b}{d}, z = \frac {c}{d}$ we will get the final answer: for $ n\in N$ \[ (x,y,z) = (\frac {2^n}{2^{4n - 3} + 1}, \frac {2^{4n - 3} - 1}{2^{4n - 3} + 1}, \frac {2^{3n - 2}}{2^{4n - 3} + 1}) \] In same way we can alsow prove, that equation $ x^4 + y^4 + 4^{2s - 1}z^4 = 1, s\in N$ has infinitely many solutions $ x,y,z\in Q$. In this case the answer will be: For $ n\geq \frac{s + 1}{3}$ \[ (x,y,z) = (\frac {2^n}{2^{4n - 3} + 1}, \frac {2^{4n - 3} - 1}{2^{4n - 3} + 1}, \frac {2^{3n - s - 1}}{2^{4n - 3} + 1}) \]