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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!

George R.
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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 $0$? – Wojowu Feb 03 '17 at 11:42

3 Answers3

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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?

Another approach, if the above is "slightly overlevel" as in the comments:

If $2^n+3^m$ were divisible by $23$ then so is $8^n+27^m$ as $a+b$ divides $a^3+b^3$. Then $8^n+4^m=2^{3n}+2^{2m}$ would also be a multiple of $23$. Through division by the smaller power of $2$ in $\{8^n,4^m\}$ we find that $2^k+1$ should be a multiple of $23$ for some whole number $k(=|3n-2m|)$. Now try to find a value of $k$ that makes $2^k\equiv-1\bmod 23$.

Oscar Lanzi
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    a very elegant solution – miracle173 Feb 03 '17 at 11:53
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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 – DonAntonio Feb 03 '17 at 12:04
  • I had to think for a second why no two nonzero squares can sum to $0$ $\pmod p$ when $p$ is a prime that is $3 \pmod 4$. But then it occured to me that if $a^2 +b^2 \equiv 0 \pmod p$ then $b^2= -a^2 \implies -1$ is a square $\pmod p$ . This is impossible for such a prime $p$, as then there would be an element $c$ such that $c^2=-1$ and $c^2=4$, which would imply by Lagrange's Thm that $4$ divides $|(\mathbb{F}_p)^*|=p-1$, which is impossible because $p-1$ is $2 \pmod 4$. – Mike Aug 02 '24 at 21:44
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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. – miracle173 Feb 03 '17 at 11:55
  • That sorting shows the sets are in fact identical. So a zero sum $\bmod 23$ would require each set of powers to contain additive inverse pairs and those would perforce include $\pm1$. We then see that $-1\equiv22$ is absent. Compare with my answer where the non-existence of the $-1$ residue emerges algebraically. – Oscar Lanzi Mar 23 '21 at 12:32
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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, ...

wendy.krieger
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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 $2^6+3^1\equiv -2$? – Wojowu Feb 03 '17 at 11:51
  • Indeed so. It turns out that all residues $\bmod 23$ can be rendered as $2^n+3^m$ -- except zero. – Oscar Lanzi Mar 23 '21 at 15:24