Is there an even number $n \in \mathbb{N}$ and two different primes $p,q<n$ which are not divisors of $n$, as well as $a,b \in \mathbb{N}$ with $a,b>1$, such that $$ n=q+p^{a}=p+q^{b} $$ ? I conjecture, there is no such even number $n$, but I'm not sure.
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3$\begingroup$ $2200=3+13^3=13+3^7$ $\endgroup$– Mike BennettCommented Feb 13, 2021 at 23:02
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$\begingroup$ Thank you. Is there theory about it or was this a bruteforce counter example? $\endgroup$– HandwavyCommented Feb 14, 2021 at 9:46
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$\begingroup$ Rearrange as $p^a-p=q^b-q$ and you get oeis.org/A057896, though not specifically primes. Maybe some useful links there. $\endgroup$– nickgardCommented Feb 14, 2021 at 10:05
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$\begingroup$ Note that for $n$ to be even, $p,q$ must have the same parity. Assuming $p\ne q$, they both must be odd. Rearrange as $p^a-q^b=p-q$ to observe that you are looking for odd primes whose difference is also the difference of some powers of each. Find those, and for each you will find an $n$ $\endgroup$– Keith BackmanCommented Feb 14, 2021 at 17:59
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$\begingroup$ @KeithBackman, since $p,q$ do not divide $n$ and $n$ is even, $p,q$ can't be even (can't be equal to 2). But I wonder if there is a systematic way to come up with those prime pairs, or maybe if one can find those exponents for every odd prime pair. $\endgroup$– HandwavyCommented Feb 15, 2021 at 13:54
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