The correct formulation requires $a$, $b$, $c$ to be positive integers (in some countries $0$ is a natural number).

Almost same as China TST 2022! (notice we must have $q=2$)

Batapan wrote: Almost same as China TST 2022! (notice we must have $q=2$) Not quite, since in the given formulation completely different (simpler) techniques suffice. Anyway, the main point was to come up with a problem which can be a close replica (in terms of looks and techniques) to JBMO 2022/3.

Assassino9931 wrote: Batapan wrote: Almost same as China TST 2022! (notice we must have $q=2$) Not quite, since in the given formulation completely different (simpler) techniques suffice. Anyway, the main point was to come up with a problem which can be a close replica (in terms of looks and techniques) to JBMO 2022/3. Ok , I see. It is a really beautiful equation anyways! So, as I said we can easily verify $q=2$. if $p=2$ or if they're both odd, we get a contradiction $mod2$ So the equation becomes $$2^bp^a=(p+4)^c+1$$Suppose $b\geq 2$. Mod $4$ we have $p^c+1 \equiv 0$ mod $4$, thus $c$ must be odd. $$2^bp^a=(p+5)(........)$$so $p+5=2^mp^n$. If $n=0$, then $p+5=2^m$ and since $p+5\not\equiv 0$ mod $16$, $m\le 3 \implies m=3$ so $p=3$, but this will give a contradiction $mod6$ If $n\geq1$ ,$modp$ we must have $p=5$ so $$2^b5^a=9^c+1$$$(1)$ which we can easily verify that $a=b=c=1$ is the only solution. If $b=1$ $$2p^a=(p+4)^c+1$$If $c$ is odd, we get back to $(1)$... Suppose $c=2r$ where $r$ is an odd positive integer. then $$2p^a=(p^2+8p+16)^r+1=(p^2+8p+17)(.......)$$so $p^2+8p+17=2p^k \implies p=17 \implies 2 \cdot 17^a=21^c+1$ Taking the last $mod16$ we get $5^c \equiv 1$ mod $16$ but by $LTE$ it is true that $u_2(5^c-1)=2+1+1-1=3$ so we reached a contradiction. Thus $(a,b,c,p,q)=(1,1,1,5,2)$

@above for $2^b5^a = 9^c +1$ arguing by moduli 4 and 25 suffices. For the final part, to avoid LTE note that $p^2+8p+17$ cannot be a power of 2 by mod 4 and hence must be divisible by $p$, inplying $p=17$ but then $2^b17^a = (p^2+8p+17)A = 442A$ is divisible by 13, contradiction.

Assassino9931 wrote: @above for $2^b5^a = 9^c +1$ arguing by moduli 4 and 25 suffices. For the final part, to avoid LTE note that $p^2+8p+17$ cannot be a power of 2 by mod 4 and hence must be divisible by $p$, inplying $p=17$ but then $2^b17^a = (p^2+8p+17)A = 442A$ is divisible by 13, contradiction. Nice

Moreover, it turned out that the condition $p\not\equiv 11 \pmod {16}$ is redundant! From the given equation we immediately have that $p$ divides the sum of two squares $4^c + 1 = 2^{2c} + 1^2$ and hence $p$ must be $1$ mod $4$.

Assassino9931 wrote: Moreover, it turned out that the condition $p\not\equiv 11 \pmod {16}$ is redundant! From the given equation we immediately have that $p$ divides the sum of two squares $4^c + 1 = 2^{2c} + 1^2$ and hence $p$ must be $1$ mod $4$. I was thinking about that