<s>[INST] Prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$? I have to prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$. 
I have proven using congruencies that $2^{15}-1$ is divided by $31$. However we have
$$2^5\equiv 10 \mod{11}$$
$$2^{15}\equiv 10^3=1000\equiv 10 \pmod {11}$$
Therefore
$$2^{15}-1\equiv 9 \pmod{11}.$$
So it is impossible to prove!!
 [/INST] You are correct: the factorization you are given is wrong.
Since $N=2^{15}-1=(2^5)^3-1=(2^5-1)((2^5)^2+2^5+1)$ you get
$$
2^{15}-1=31\cdot(1024+32+1)=31\cdot 1057
$$
However, you can also use
$$
2^{15}-1=(2^3)^5-1=(2^3-1)((2^3)^4+(2^3)^2+2^3+1)
$$
so $N$ is also divisible by $7$. Divide $1057$ by $7$ to get
$$
N=31\cdot 7\cdot 151
$$
(and $151$ is prime).
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