Is
$$13^2 + 7^3 = 512=2^9$$
the only solution for the sum of two primes $p,q$ raised to powers greater than $1$ equals a third prime power?
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Well, there is $2^2+2^2=2^3$. – Oscar Lanzi Dec 16 '18 at 21:53
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2\begin{eqnarray} 2^5+7^2=3^4 \end{eqnarray} There are other nice examples here ... http://people.math.sfu.ca/~ichen/pub/BeCh2.pdf – Donald Splutterwit Dec 16 '18 at 21:55
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1Since $1+1 \not \equiv 1, \rm{mod}, 2$, the problem is equivalent to find more solutions to $2^a = p^b \pm q^c$. – Lucas Henrique Dec 16 '18 at 21:57
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No, it isn't. For instance, we have:
- $2^n+2^n=2^{n+1}$, for all $n$ meeting problem constraints; and
- $2^4+3^2=5^2$.
Oscar Lanzi
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Aside from the one with all 2s, it appears that my specimen is the only one to have the sum equal a power of 2. I checked that PDF and NO other with sum being a power of 2. – J. M. Bergot Dec 16 '18 at 22:16
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1The pdf answered my question. Your site has been very helpful. Do keep up the excellent assistance to the laity. – J. M. Bergot Dec 18 '18 at 18:43
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You can check these types of numbers on wolfram alpha.
However you can see that $3^1+6^1=9^1$
Tianlalu
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It is fair to say that 3 and 6 are not odd primes to a power >1 and that 9 is not a power of 2. – J. M. Bergot Dec 19 '18 at 20:06