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Find the smallest positive integer $n$ such that $125|2^n+3^n$

Suppose $125|2^n+3^n$. Now, $n$ must be odd because if $n$ is even then $$0\equiv 2^n+3^n\equiv2^n+(-2)^n\equiv 2^n+2^n\equiv2^{n+1}\pmod 5,$$
which is in contrast to hypothesis. How to go on from here?

Math Lover
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Hamid Reza Ebrahimi
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  • $n=25$ is the smallest $n$ doing the job. But I found this answer with PARI/GP. Probably you want a solution by hand. – Peter Jan 01 '18 at 15:27
  • If you look at powers of $2$ and $3$ modulo $25$, it's clear that we must have $n\equiv 5\pmod{10}$. That narrows your search considerably. – G Tony Jacobs Jan 01 '18 at 15:33

4 Answers4

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Lifting the exponent lemma is useful here.

Let $p>2$ be a prime. Let $a$ and $b$ be integers not divisible by $p$ and let $n$ be an odd positive integer. If $p\mid a+b$ then $$v_p(a^n+b^n)=v_p(a+b)+v_p(n).$$ Here $v_p(x)$ denotes the unique integer with the property that $p^{v_p(x)} \mid x$ and $p^{v_p(x)+1} \nmid x$.

Using the lemma for $p=5$, $a=2$, and $b=3$ we find that the smallest $n$ is $25$.

timon92
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Using Proof of $a^n+b^n$ divisible by a+b when n is odd,

$$2^n+3^n$$ will be divisible by $2+3$ iff $n$ is odd $=2m+1$(say)

$$2^{2m+1}+3^{2m+1}=2^{2m+1}+(5-2)^{2m+1}$$

$$\equiv\binom{2m+1}15\cdot2^{2m}-\binom{2m+1}25^2\cdot2^{2m-1}\pmod{5^3}$$

So, we need

$$25\mid2^{2m-1}(2m+1)(2-5m)$$

$$\iff25\mid(2m+1)$$ as $5\nmid2^{2m-1}(2-5m)$

lab bhattacharjee
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Certainly $2^{25}+3^{25}$ is divisible by $125$. One may check now by computation that $n=25$ is the smallest such $n$.

Dietrich Burde
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As $3\equiv2^7\pmod{125}$

$$2^n+3^n\equiv2^n+(2^7)^n\equiv2^n(2^{6n}+1)\pmod{125}$$

So, we need $2^{6n}\equiv-1\pmod{125}$

Using Show that $r$ is a primitive root (mod $p^k$), $2$ is a primitive root $5^n$ for $n\ge1$

$$\implies6n\equiv\dfrac{\phi(125)}2\pmod{\phi(125)}$$

$$\iff3n\equiv25\pmod{50}$$

As $17\cdot3\equiv1\pmod{50}$

$$ n\equiv25\cdot17\pmod{50}\equiv25$$

lab bhattacharjee
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