## [Soln] 2016 Japan Math Olympiad Preliminary Problem 1

2016 JMO Prelim 1. Calculate the value of $\sqrt{\displaystyle\frac{11^4 + 100^4 + 111^4}{2}}$ and answer in the form of an integer.

Two obvious facts that help get us started:

1. That the answer is an integer should tell us that the expression inside the square root is a perfect square.
2. $11 + 100 = 111$. Perhaps we should generalise and see if $\displaystyle\frac{a^4 + b^4 + (a+b)^4}{2}$ is a perfect square.

\begin{aligned} a^4 + b^4 + (a+b)^4 &= 2a^4 + 2b^4 + 4a^3b + 6a^2b^2 + 4ab^3, \\ \displaystyle\frac{a^4 + b^4 + (a+b)^4}{2}&= a^4 + b^4 + 2a^3b + 2ab^3 + 3 a^2b^2. \end{aligned}

Let’s try to match this to a perfect square expression. The $a^4$ and $b^4$ terms suggest that there should be $a^2$ and $b^2$ terms in the square, i.e.:

$a^4 + b^4 + 2a^3b + 2ab^3 + 3a^2b^2 = (a^2 + b^2 + X)^2$

for some $X$. Let’s expand the above and see what $X$ could be:

\begin{aligned} a^4 + b^4 + 2a^3b + 2ab^3 + 3a^2b^2 &= a^4 + b^4 + X^2 + 2a^2b^2 + 2a^2X + 2b^2X, \\ 2a^3b + 2ab^3 + a^2b^2 &= X^2 + 2a^2X + 2b^2X, \\ ab(ab + 2a^2 + 2b^2) &= X(X + 2a^2 + 2b^2). \end{aligned}

It’s clear that $X = ab$ fulfills the above, i.e. for any $a$ and $b$,

$\displaystyle\frac{a^4 + b^4 + (a+b)^4}{2} = (a^2 + b^2 + ab)^2.$

Substituting $a = 11$ and $b = 10$, we have

\begin{aligned} \sqrt{\displaystyle\frac{11^4 + 100^4 + 111^4}{2}} &= \sqrt{(11^2 + 100^2 + 11 \cdot 100)^2} \\ &= 121 + 10000 + 1100 \\ &= \mathbf{11221}.\end{aligned}