# Find the positive integers $m,n$ such that $2^{m}\cdot3^{n}-1$ is a perfect square.

Find the positive integers $m,n$ such that $2^{m}\cdot3^{n}-1$ is a perfect square.

I tried to let $2^{m}\cdot3^{n}-1=x^2$ then $x$ is an odd number but here i don't have any ideas. I think it can use Fermat-Euler theorem. Is it a right way? How can i solve it?

• If $n$ is positive then your expression is congruent to $2$ mod $3$, which cannot be a square – TomGrubb Jul 12 '17 at 16:59
• Hello and welcome to math.stackexchange. Interesting question - where does it come from? Have you tried some examples? – Hans Engler Jul 12 '17 at 16:59

Since $m,n$ are positive then the number $2^m3^n-1$ will be of the form $6k-1$ or $6k+5$. Now notice that no square is of the form $6k+5$.

# First claim: $n \leq 0$.

Suppose on contraary that $1 \leq n$, so then we have:

$2^m3^n-1\overset{3}{\equiv}0-1\overset{3}{\equiv}-1$,

which contradicts with the fact that, the only possible values for squares module $3$, are $0$ and $1$.

# Socend claim: $m \leq 1$

Suppose on contraary that $2 \leq m$, so then we have:

$2^m3^n-1\overset{4}{\equiv}0-1\overset{4}{\equiv}-1$,

which contradicts with the fact that, the only possible values for squares module $4$, are $0$ and $1$.

So the only possibilities for $2^m3^n-1$ to be square, are the below cases:

$n=0, m=0$;

$n=0, m=1$.

The first case gives us $x=0$, and the second gives $x=1$.

• Looks good. But did you accidentally write the inequalities the opposite way? – Jyrki Lahtonen Jul 12 '17 at 22:01

Let's assume $$2^m3^n-1=x^2 \tag{1}$$ If we assume $$n>0$$ then $$\gcd(x, 3)=1$$ (otherwise $$3 | 1$$), thus, from Euler's theorem $$x^2 \equiv 1 \pmod{3}$$ but from $$(1)$$: $$x^2 \equiv -1 \pmod{3} \Rightarrow 0\equiv 2 \pmod{3}$$ which is a contradiction, so $$n=0$$.

Now we look at $$2^m-1=x^2 \tag{2}$$ and assume $$m\geq1$$. As you already mentioned, $$x$$ must be odd, e.g. $$x=2q+1$$, then $$2^m-1=(2q+1)^2 \Rightarrow 2^{m-1}=2q(q+1)+1$$ which limits $$m$$ to $$1$$.

Altogether $$m \in \{0,1\}, n=0$$.