$4k^4+4k^2=(2k^2)^2+(2k)^2$, $4k^4+4k^2+1=(2k^2+1)^2+0^2$, $4k^4+4k^2+2=(2k^2+1)^2+1^2$ $\forall$ $k\in\mathbb{Z}$

Notice that $8=4^2+4^2$, $9=3^2+0^2$, $10=3^2+1^2$. Now, assume we have a triplet $(a,a+1,a+2)$, such that: $$a=x^2+y^2\quad a+2=z^2+t^2$$Then notice that $a^2+2a=a(a+2)=(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2$ (this is known as Lagrange's identity) and $a^2+2a+1=(a+1)^2+0^2$, $a^2+2a+2=(a+1)^2+1^2$, so since we start with the triplet for $a=8$, obviously $a^2+2a>a$ therefore we cen generate infinitely many "working" triplets of integers by $(a,a+1,a+2)\rightarrow (a^2+2a,a^2+2a+1,a^2+2a+2)$ repeated infinitely many times, so we're done.