IMO contest problem (Second solution) Prove that $a^2+b^4=1994$

Let $a,\,b$ be positive integers with $b\gt3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$.

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

The solution below is provided by the retired math professor from the U.K. In my opinion, this solution is more easy to follow than the previously discussed, but credit must be given to both as they are equally insightful and neat solutions.

If $a^2+b^4=2\bigl((a-6)^2+(b+1)^2\bigr) = 2a^2 + 2b^2 - 24a + 4b + 74$ then $b^4 - 2b^2 - a^2 + 24a = 4b+74.$ Complete the squares on the left, to get $(b^2-1)^2 - (a-12)^2 = 4b-69.$

The two squares on the left cannot be equal, because that would mean $4b-69=0$, and that does not have an integer solution. So there are two possible cases.

Case 1: $b^2-1 \gt a-12$.

In this case, $a-12$ is at most $b^2-2$.

Therefore $4b-69 = (b^2-1)^2 - (a-12)^2 \geqslant (b^2-1)^2 - (b^2-2)^2 = 2b^2-3.$ But this says that $0\geqslant 2b^2 - 4b + 66 = 2(b-1)^2 + 64.$ That is clearly impossible, so this case cannot arise.

Case 2: $b^2-1 \lt a-12$.

Then $a-12$ is at least $b^2$, and so $4b-69 =(b^2-1)^2 - (a-12)^2 \leqslant (b^2-1)^2 - (b^2)^2 = 1-2b^2.$ Therefore $2b^2 + 4b - 70 \leqslant0.$ This says that $2(b+1)^2 \leqslant 72$, so that $(b+1)^2 \leqslant 36$, $b+1 \leqslant 6$, $b\leqslant5.$ But we are told that $b>3$. Therefore $b=4$ or $5$.

If $b=4$ then the equation $(b^2-1)^2 - (a-12)^2 = 4b-69$ becomes $(a-12)^2 = 278.$ But that is not a perfect square, so we must have $b=5$, in which case $(a-12)^2 = 625 = 25^2$, and $a = 25+12 = 37.$

Finally, $a^2+b^4 = 37^2 + 5^4 = 1369 + 625 = 1994.$