2012 Indonesia MO Problems/Problem 7

Problem

Let $n$ be a positive integer. Show that the equation $\sqrt{x}+\sqrt{y}=\sqrt{n}$ have solution of pairs of positive integers $(x,y)$ if and only if $n$ is divisible by some perfect square greater than $1$ .

Solution

Since iff is a double implication, we can prove that if there exists a positive integer solution $(x,y)$ to $\sqrt{x}+\sqrt{y}=\sqrt{n}$ , then $n$ is divisible by some perfect square greater than $1$ , and if $n$ is divisible by some perfect square greater than $1$ then there exists a positive integer solution (x,y) for $\sqrt{x}+\sqrt{y}=\sqrt{n}$ .

Lets tackle the latter first, let $n=m^2p$ where $m>1$ and $p$ is not divisible by any perfect square greater than $1$ , let $x=p(m-1)^2$ and $y=p(1)^2$ . Substituting back in we can get $\sqrt{p(m-1)^2}+\sqrt{p(1)^2}=\sqrt{m^2p}\implies (m-1)\sqrt{p}+\sqrt{p}=m\sqrt{p}\implies m\sqrt{p}=m\sqrt{p}$ which is true, thus it is proven

For the first, let $x=a^2b$ and $y=c^2d$ where $b,d$ are not divisible by a perfect square greater than $1$ , $\sqrt{x}+\sqrt{y}=\sqrt{n}\implies x+y+2\sqrt{xy}=n$ . Since $\sqrt{xy}$ has to be an integer, then $xy$ must be a perfect square, that means $a^2c^2bd$ is a perfect square which means $bd$ is a percect square, let $b=p_1p_2\dots p_i$ where $p$ are distinct primes, for $bd$ to be a perfect square, $d$ must be exactly $p_1p_2\dots p_i$ , as if it were less there exists a $p_k$ that divides $b$ but not $d$ and thus would not be a perfect square, the same logic would apply if $d$ was bigger than $b$ , thus $b=d$ . $x+y+2\sqrt{xy}=n$ $a^2b+c^2b+2\sqrt{a^2b^2c^2}=n$ $a^2b+c^2b+2abc=n$ $b(a^2+c^2+2ac)=n$ $b(a+c)^2=n$ since $a,c\geq 1\implies a+c\geq 2$ , thus n is divisible by a perfect square greater than 1

2012 Indonesia MO (Problems)
Preceded by Problem 6	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 8
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2012 Indonesia MO Problems/Problem 7

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