A question about perfect powers

Question

For what basis $b>6$ the number $5654$ is a power of a prime number? I posted my proof but i want to know if there are other proofs( for example though congruences). What do you think?

Answer 1

From $$p^m=N=(b+1)(5b^2+b+4)$$ and $5b^2+b+4>b+1$ we conclude $b+1\mid 5b^2+b+4$. But then also $5b^2+b+4=(b+1)(5b-4)+8$, so that $b+1\mid 8$. So $b+1\in\{1,2,4,8\}$ and together with $b>6$, we obtain $b=7$ as only possibility (in which case indeed $N=2^{11}$).

Answer 2

$$N=5b^3+6b^2+5b+4=5b^3+5b^2+b^2+4b+b+4=(b+1)(5b^2+b+4)$$ We can not that if $b$ is odd: $(b+1)$ is even, while if $b$ is even $(5b^2+b+4)$ is even, therefore $N$ is always even and is a power of two. $(b+1)$ and $(5b^2+b+4)$ are both powers of two and therefore $(b+1)= 2^n$ and $(5b^2+b+4)=2^m$. Replacing in the second equation we can write $$5\cdot 2^{2n}-9\cdot 2^n+8=2^m$$ and dividing for $8$ we obtain:$$5\cdot 2^{2n-3}-2^{m-3}=9\cdot 2^{n-3}-1$$ $5\cdot 2^{2n-3}-2^{m-3}$ is even also $9\cdot 2^{n-3}-1$ is even and therefore $n=3$. The equation becomes $5\cdot 8 -9+1=2^{m-3}$ and $m=8$. The only solution is b=7.