Friday, June 19, 2015

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?

There are two given conditions that are the main guidance that could save us time and effort to solve the problem.

We have to always bear in mind that

1. Both $a$ and $b$ are positive integers, and

2.  $b>3$.

We see that the equation, after expansion, consists of two parts that define by the variables, $a$ and $b$:





Rewriting the given equation we have, a quadratic in $a$ where


It has integer solutions only if the discriminant is a perfect square.


Note that for $b\ge 4$,





which implies [MATH]\color{yellow}\bbox[5px,blue]{(b^2-2)^2\lt b^4-2b^2-4b+7}[/MATH].





which implies [MATH]\color{yellow}\bbox[5px,blue]{b^4-2b^2-4b+7\lt (b^2+1)^2}[/MATH].

We can conclude by now that

$(b^2-2)^2\lt b^4-2b^2-4b+7\lt (b^2+1)^2$

The only perfect squares between $(b^2-2)^2$ and $(b^2-1)^2$ are $(b^2-1)^2$ and $(b^2)^2$.

When $b^4-2b^2-4b+70=(b^2-1)^2$.

That gives $b=\dfrac{69}{4}$ which isn't an integer.

When $b^4-2b^2-4b+70=(b^2)^2$, that gives


$b=5$ or $b=-7$.

Thus, $b=5$ and it gives $a=37$ and $a^2+b^4=37^2+5^4=1994$.

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