Let $a,\,b$ be positive integers with $b>3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$. Prove that $a^2+b^4=1994$.

Let $a,\,b$ be positive integers with $b>3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$.

Prove that $a^2+b^4=1994$.

This is a genuinely hard problem. If you are smart, you will wonder and ask your teacher if this question is saying there is only one solution for $(a,\,b)$ such that there is only a value for $a^2+b^4$, and that must be $1994$.

Yes, your intuition is right, the question is, how to find that only solution for both $a$ and $b$ from the only given single equation?