abdurashidjon wrote:

grobber wrote:

Assume $n$ is not a power of $2$ . In this case there is an odd number $k\ge 3$ s.t. $2^k-1|2^n-1|m^2+9$ . $2^k-1$ is of the form $4k+3$ , so it has a prime factor of the form $4t+3$ and $\ne 3$ , because $3\not |\ 2^k-1$ , so it can't divide $m^2+9$ because if a prime of the form $4t+3$ divides $a^2+b^2$ , then it divides $a,b$ .

Hi here grobber has used that if an number is of the form $4k+3$ then it will a prime divisor of the same form, and also, if a number of the form $4k+3$ divides $a^2+b^2$ then it have to divide $a, b$
Abdurashid

Seven years ago, but for future reference, here is the proof. Consider a number of the form $4k+3.$ Clearly, all its prime factors must be of the form $4m+1$ or $4m+3.$ Since $4k+3 \equiv 3\pmod 4,$ the factors cannot all be $1\pmod 4,$ so there must be at least one prime factor of the form $4m+3.$

Now suppose that a prime $4m+3$ divides $a^2+b^2.$ Let $a^2 \equiv j\pmod{4m+3}.$ Assume for sake of contradiction that $j \neq 0.$ Then $b^2 \equiv -j\pmod{4m+3}.$ So we obtain $a^{4m+2} \equiv j^{2m+1}\pmod{4m+3}$ and $b^{4m+2} \equiv -j^{2m+1}\pmod{4m+3}.$ However, by FLT, we must have $j^{2m+1}=-j^{2m+1}=1,$ a contradiction. Therefore, $a \equiv b \equiv 0\pmod{4m+3}.$