$n$ and $m$ are integers and $m >1$. Given: $2^n −2=m(m+1)$, Prove that $n$ can’t be an even number .

Question

$n$ and $m$ are integers and $m >1$ we have : $2^n −2=m(m+1).$

Prove that $n$ can’t be an even number .

Answer 1

Let $n=2k$ and multiply the equation with 4:

$$4^{k+1}-7 = 4m^2+4m+1 = (2m+1)^2$$

So $$\Big(2^{k+1}-(2m+1)\Big) \Big(2^{k+1}+(2m+1)\Big) =7$$

Then $$2^{k+1}+(2m+1) =7$$ $$2^{k+1}-(2m+1) =1$$

If we substract them we get: $$4m+2=6\implies m=1 \implies 2^n = 4\implies \boxed{n=2}$$ But this is not true since $m>1$.

Answer 2

Let $m\equiv 0,1,2,-1\pmod 4$ then $m(m+1) \equiv 0,2,2,0\pmod 4$.

If $n=2k$ is even (and assuming as $m(m+1) > 2$ that $k\ge 1$) then $2^n -2\equiv 2\pmod 4$

If $m \equiv 1 \pmod 4$ we have $m = 4j+1$ so

$2^{2k} - 2 = (4j+1)(4j+2)=16j^2 +12j +2$ and $2^{2k} = 16j^2 + 12j + 4$ and

$2^{2k-2} = 4j^2+3j + 1$ which is odd. So $k=1$ and $j=0$ so $m =1$.

If $m \equiv 2\pmod 4$ so $m = 4j+2$ we have

$2^{2k} -2 = (4j+2)(4j+3) = 16j^2 + 20j + 6$ so

$2^{2k} = 16j^2 + 20j + 8$ so

$2^{2k-2} = 4j^2 + 5j + 2$.

Now if $2k-2=0$ we have $4j^2 + 5j +1=0$ so $j=-1$ and $m = -2$.

Now if $2k-2>0$ we must have $4|5j+2$ so $j\equiv 2\pmod 4$ and so $j$ is even but not divisible by $4$..

$2^{2k-3} = 2j^2 + 5\frac j2 + 1\equiv 2\pmod 4$.

So $2^{2k-3} = 2$. and $2k-3 =1$ which isn't possible.