THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

We treat four cases:

I. $x,y\ge 0$

$\sqrt{4^x+5^y}$ is rational if and only if $4^x+5^y$ is a perfect square, i.e., there exists $n\in \mathbb{N}$ such that $4^x+5^y=n^2$. Analyzing this equation modulo $3$, we have $4^x\equiv 1(\mathrm{mod}3)$, $5^y\equiv (-1{)}^y(\mathrm{mod}3)$, and $n^2\equiv 0,1(\mathrm{mod}3)$, hence $y$ needs to be odd. Let $z\in \mathbb{N}$ be such that $y=2z+1$. The previous equation becomes $4^x+5\cdot 25^z=n^2$.

If $x\ge 2$, then $4^x\equiv 0(\mathrm{mod}8)$, $5\cdot 25^z\equiv 5(\mathrm{mod}8)$, while $n^2\equiv 0,1,4(\mathrm{mod}8)$, which means the equation has no solutions in this case. We are left with the cases when $x=0$ and $x=1$.

For $x=0$ we have $4^x+5^y\equiv 2(\mathrm{mod}4)$, hence $4^x+5^y$ cannot be a perfect square.

If $x=1$, then $5^y=(n-2)(n+2)$, hence there exist $a,b\in \mathbb{N}$, with $a+b=y$, such that $n-2=5^a$ and $n+2=5^b$. By subtraction, $5^b-5^a=4$. If $a,b\ge 1$ then $5\mid 5^b-5^a$, hence $5\mid 4$, contradiction. As $a<b$, it follows that $a=0$, then $b=1$, i.e., $y=1$. We obtain the solution $x=y=1$.

II. $x,y<0$

Let $u=-x,v=-y,u,v>0$. $\sqrt{4^x+5^y}$ is rational if and only if there exist $p,q\in {\mathbb{N}}^{\ast }$, coprime, such that $4^x+5^y=\frac{p^2}{q^2}$, i.e., $\frac{4^u+5^v}{4^u\cdot 5^v}=\frac{p^2}{q^2}$, which means $p^2\cdot 4^u\cdot 5^v=q^2(4^u+5^v)$. Numbers $4^u\cdot 5^v$ and $4^u+5^v$ are coprime, therefore every prime factor of $4^u+5^v$ is a prime factor of $p^2$, which means that it appears at an even exponent. It follows that $4^u+5^v$ is a perfect square, and, similarly, $4^u\cdot 5^v$ is a perfect square. From case I it follows that $4^u+5^v$ is a perfect square if and only if $u=v=1$, but then $4^u\cdot 5^v=20$ is not a perfect square. We conclude that there are no solutions in this case.

III. $x<0,y\ge 0$

Let $u=-x\in {\mathbb{N}}^{\ast }$. Then $\sqrt{4^x+5^y}=\frac{\sqrt{1+4^u\cdot 5^y}}{2^u}$ is rational if and only if $1+4^u\cdot 5^y$ is a perfect square, i.e., there exists $n\in \mathbb{N}$ such that $1+4^u\cdot 5^y=n^2$. Then $4^u\cdot 5^y=(n-1)(n+1)$. As $u>0$, $n$ is odd and $(n-1,n+1)=2$. We distinguish the following sub-cases:

A. $n-1=2\cdot 5^y$, $n+1=2^{2u-1}$

B. $n-1=2$, $n+1=2^{2u-1}\cdot 5^y$

C. $n-1=2^{2u-1}$, $n+1=2\cdot 5^y$

D. $n-1=2^{2u-1}\cdot 5^y$, $n+1=2$

In sub-case A we obtain $2^{2u-2}-5^y=1$, i.e., $(2^{u-1}-1)(2^{u-1}+1)=5^y$. It follows that $2^{u-1}-1$ and $2^{u-1}+1$ should be powers of $5$, but no two powers of $5$ are at distance $2$.

Sub-case B leads to $n=3$ and, immediately, to $4=2^{2u-1}\cdot 5^y$, with no solutions.

In sub-case C we get $5^y-2^{2u-2}=1$. Then $5^y\equiv (-1{)}^y(\mathrm{mod}3)$ and $2^{2u-2}=4^{u-1}\equiv 1(\mathrm{mod}3)$, therefore $y$ needs to be odd. It follows that $5^y\equiv 5(\mathrm{mod}8)$, hence $2^{2u-2}\equiv 4(\mathrm{mod}8)$, i.e., $u=2$. We obtain the solution $x=-2,y=1$.

Sub-case D leads to $n=1$ and then to $0=2^{2u-1}\cdot 5^y$, with no solutions.

IV. $x\ge 0,y<0$

Let $v=-y\in {\mathbb{N}}^{\ast }$. Then $\sqrt{4^x+5^y}=\sqrt{\frac{1+4^x\cdot 5^v}{5^v}}$ is rational if and only if there exist $p,q\in {\mathbb{N}}^{\ast }$ coprime such that $\frac{1+4^x\cdot 5^v}{5^v}=\frac{p^2}{q^2}$, i.e., such that $p^2\cdot 5^v=q^2(1+4^x\cdot 5^v)$. Numbers $1+4^x\cdot 5^v$ and $5^v$ are coprime, hence, as in case I, they need to be perfect squares. We have seen in the previous case that $1+4^x\cdot 5^v$ is a perfect square only when $x=2$ and $v=1$, but in this situation $5^v=5$ is not a perfect square. We conclude that in this case there are no solutions.

To summarize, the only solutions to the problem are $x=y=1$ and $x=-2$, $y=1$.