2012 Indonesia MO Problems/Problem 7

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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) fo$\sqrt{x}+\sqrt{y}=\sqrt{n}$.

Lets tackle the latter first, let$ (Error compiling LaTeX. Unknown error_msg)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$\sqrt{p(m-1)^2}+\sqrt{p(1)^2}=\sqrt{m^2p}\implies (m-1)\sqrt{p€+\aqrt{p}=m\sqrt{p}\implies m\sqrt{p}=m\sqrt{p}$which is true, thus it is proven

For the first, let$ (Error compiling LaTeX. Unknown error_msg)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$b=d$. <cmath>x+y+2\sqrt{xy}=n</cmath> <cmath>a^2b+c^2b+2\sqrt{a^2b^2c^2=n</cmath> <cmath>a^2b+c^2b+2abc=n</cmath> <cmath>b(a^2+c^2+2ac)=n</cmath> <cmath>b(a+c)^2=n</cmath> since$ (Error compiling LaTeX. Unknown error_msg)a,c\geq 1\implies a+c\geq 2$, thus n is divisible by a perfect square greater than 1

See Also

2012 Indonesia MO (Problems)
Preceded by
Problem 6
1 2 3 4 5 6 7 8 Followed by
Problem 8
All Indonesia MO Problems and Solutions