AlastorMoody wrote:
Solve the given equation in primes
\begin{align*} xyz=1 +2^{y^2+1} \end{align*}
Are you sure this is supposed to be in primes? Coz then we get $$2^2 \equiv 2^{y^2+1} \equiv -1 \pmod{y} \Rightarrow y=5$$One can easily check (using Google ) that $2^{26}+1$ has 4 prime factors, namely $5,53,157,1613$. So there are no solutions.
EDIT: Confirmed in PM that this is indeed in primes
@below $2^{y^2} \neq (2^y)^2$
@above: maybe typo?
@above: HAHAHA, my bad what's in my brain?, may I fix it?
$$2^{y^2+1}=(2^y)^2\cdot 2\equiv 2^3\equiv -1\mod y $$
So $y=3$. Now we have $2^{3^2+1}-1=1023=3\cdot 11\cdot 31$. This implies $(x,y,z)=(11,3,31) $.