2
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ If $n$ is positive then your expression is congruent to $2$ mod $3$, which cannot be a square $\endgroup$ – TomGrubb Jul 12 '17 at 16:59
  • $\begingroup$ Hello and welcome to math.stackexchange. Interesting question - where does it come from? Have you tried some examples? $\endgroup$ – Hans Engler Jul 12 '17 at 16:59
2
$\begingroup$

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 $6k+5$.

$\endgroup$
1
$\begingroup$

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 $x=1$.

$\endgroup$
1
  • $\begingroup$ Looks good. But did you accidentally write the inequalities the opposite way? $\endgroup$ – Jyrki Lahtonen Jul 12 '17 at 22:01
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.