I want to prove or disprove that for a natural number $n$, if $10|n^4$, then $10|n$. I'm really struggling to decide on how to comprehensively prove it or not, because all of the other related questions I've found on this site seem predicated on using Euclid's Lemma, which necessitates the divisor being prime.
From what I've been reading and learning, I think the fact that $10$ is a product of unique primes is important to proving the statement, but I don't understand how enough to be able to confidently assert anything. I'm thinking that it's true, because the smallest fourth power of a natural number that is a multiple of $10$ is $10,000$, or $10^4$, and obviously $10|10$— and it seems moving further $10$ only divides multiples of itself raised to the fourth power. However, I'm struggling to see how I can put that into an appropriate explanation.
Best Answer
If $10|n^4$ then $2|n^4$ and $5|n^4$. Since $2$ and $5$ are prime numbers we must have $2|n$ and $5|n$ (because if a prime divides a product $ab$ then it must divide either $a$ or $b$). But that implies that $lcm(2,5)|n$. Since $2$ and $5$ are relatively prime their lcm is their product. So indeed $10|n$.