Show that $\frac{a^4+b^4+(a+b)^4}{2}$ is a perfect square.
I tried this, $$\frac{a^4+b^4+(a+b)^4}{2}$$ $$\frac{a^4+b^4+(a^2+b^2+2ab)^2}{2}$$ $$\frac{2a^4+2b^4+4a^2b^2+2(a^2b^2+2a^3b+2ab^3)}{2}$$ $$a^4+b^4+2a^2b^2+ab(ab+2a^2+2b^2)$$ $$(a^2+b^2)^2+ab(ab+2a^2+2b^2)$$
What can I do next?
If $a^4+2a^3b+3a^2b^2+2ab^3+b^4$ is a square, it has to be of the form $(xa^2+yab+zb^2)^2$ for some coefficients $x,y,z$.
Then $a^4+2a^3b+3a^2b^2+2ab^3+b^4=x^2a^4+2xya^3b+(y^2+2xz)a^2b^2+2yzab^3+z^2b^4$.
Comparing coefficients we get $x^2=1, 2xy=2, y^2+2xz=3,2yz=2,z^2=1$.
Solving this system we get $x=y=z=1$ and $x=y=z=-1$, so $a^4+2a^3b+3a^2b^2+2ab^3+b^4=(a^2+ab+b^2)^2$