$x^2+yz=u^4 \Rightarrow (u^2+x)(u^2-x)=yz \Rightarrow u^2-x=1 $ and $ u^2+x=yz, $ then we get by mod 3 that $ x=3,u=2, $ but then $ yz=7 $ which is impossible, or $u^2-x=y $ and $ u^2+x=z$ $\Rightarrow x^2+yu<x^2+u^3<(x+u)^3$ $\Rightarrow v=2, x^2+yu=x^2+u^2+2xu$ $ \Rightarrow y=u+2x \Rightarrow 3x=u(u-1) \Rightarrow u=3, x=2 \Rightarrow y=7$ and $ z=11, $ and the last case $ u^2+x=y $ and $ u^2-x=z \Rightarrow x^2+yu=x^2+u^3+xu<(x+u)^3 \Rightarrow v=2 \Rightarrow u^2=u+x \Rightarrow u(u-1)=x \Rightarrow u=2$ and $ x=2 \Rightarrow y=6$ which is impossible.

iDra36 wrote: $x^2+yz=u^4 \Rightarrow (u^2+x)(u^2-x)=yz \Rightarrow u^2-x=1 $ and $ u^2+x=yz, $ then we get by mod 3 that $ x=3,u=2, $ but then $ yz=7 $ which is impossible, or $u^2-x=y $ and $ u^2+x=z$ $\Rightarrow x^2+yu<x^2+u^3<(x+u)^3$ $\Rightarrow v=2, x^2+yu=x^2+u^2+2xu$ $ \Rightarrow y=u+2x \Rightarrow 3x=u(u-1) \Rightarrow u=3, x=2 \Rightarrow y=7$ and $ z=11, $ and the last case $ u^2+x=y $ and $ u^2-x=z \Rightarrow x^2+yu=x^2+u^3+xu<(x+u)^3 \Rightarrow v=2 \Rightarrow u^2=u+x \Rightarrow u(u-1)=x \Rightarrow u=2$ and $ x=2 \Rightarrow y=6$ which is impossible. Isn’t x=2, y=7, z=11, u=3, and v=2 a solution though?

CircleGeometryGang wrote: iDra36 wrote: $x^2+yz=u^4 \Rightarrow (u^2+x)(u^2-x)=yz \Rightarrow u^2-x=1 $ and $ u^2+x=yz, $ then we get by mod 3 that $ x=3,u=2, $ but then $ yz=7 $ which is impossible, or $u^2-x=y $ and $ u^2+x=z$ $\Rightarrow x^2+yu<x^2+u^3<(x+u)^3$ $\Rightarrow v=2, x^2+yu=x^2+u^2+2xu$ $ \Rightarrow y=u+2x \Rightarrow 3x=u(u-1) \Rightarrow u=3, x=2 \Rightarrow y=7$ and $ z=11, $ and the last case $ u^2+x=y $ and $ u^2-x=z \Rightarrow x^2+yu=x^2+u^3+xu<(x+u)^3 \Rightarrow v=2 \Rightarrow u^2=u+x \Rightarrow u(u-1)=x \Rightarrow u=2$ and $ x=2 \Rightarrow y=6$ which is impossible. Isn’t x=2, y=7, z=11, u=3, and v=2 a solution though? Yea he mentioned that