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Prove that $23$ does not divide $2^n + 3^m$, for any $m, n \in \mathbb{N}$.

I've tried to prove that the equation $2^n + 3^m = 0$ has no solutions in $\mathbb{Z}_{23}$, but didn't succed.

Thank you!

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    Hint: the residues of powers of $2$ modulo $23$ are $1,2,4,8,16,9,18,13,3,6,12$. What are the residues of powers of $3$? Do any two sum to $0$? – 2017-02-03

3 Answers 3

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There are only 11 modului that $2^n$ and $3^n$ can be, and the other remaining 11 modulii are the 11 numbers required to add be added to give 23.

The fact that 23 is 3, mod 4, but is -1, mod 24, means that if $2^m+3^n$ gives a value, mod 23, then the negative is not in that this form,

The same applies to any prime that gives 23 mod 24, such as 47, 71, 167, ...

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    "if $2^m+3^n$ gives a value mod 23, then the negative is not of that form" How about $2^0+3^0\equiv 2$ and $2^6+3^1\equiv -2$? – 2017-02-03
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Hint: Render $2\equiv 5^2$, $3\equiv 7^2$ mod $23$. Can a sum of nonzero squares have zero residue mod $23$ when $23$ is a $4k+3$ prime?

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    a very elegant solution – 2017-02-03
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    Very nice, though perhaps slightly overlevel for such a basic question. Anyway, I think the OP (and we all, in fact) could learn from this. +1 – 2017-02-03
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The residues $\mod23$ of the sequences $(2^n)_n$ and $(3^n)_n$ are, respectively,

$$ D=\{2^n\mod23:n\in\mathbb{N}\}=\{1,2,4,8,16,9,18,13,3,6,12\} $$

$$ T=\{3^m\mod23:m\in\mathbb{N}\}=\{1,3,9,4,12,13,16,2,6,18,8\}. $$

You need to show that $\{23\}\not\in D+T$, which can be easily accomplished by observing that whenever $a$ is in $D$, $23-a$ is not in $T$.

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    a simple but efficient procedure. If one displays the elements of the sets as a sorted sequence it is easier to check this. – 2017-02-03