# solve in N²

solve in N²:
m² = 1+2ⁿ

#### SlipEternal

MHF Helper
$$m^2-1 = 2^n$$

$$(m+1)(m-1) = 2^n$$

This implies that both $m-1$ and $m+1$ are both powers of $2$. However, there are no distinct powers of 2 that are a distance of 2 apart, so there are no solutions over the natural numbers.

1 person

#### topsquark

Forum Staff
$$m^2-1 = 2^n$$

$$(m+1)(m-1) = 2^n$$

This implies that both $m-1$ and $m+1$ are both powers of $2$. However, there are no distinct powers of 2 that are a distance of 2 apart, so there are no solutions over the natural numbers.
m = 3, n = 3.

-Dan

1 person

MHF Helper