# Prove or disprove: if $10$ divides $n^{4}$, then $10$ divides $n$

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.

• If $n$ "ends" in a nonzero digit, then so does $n^2$, and also $(n^2)^2$. Mar 9 '19 at 2:16

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 $$10|n$$.

Yes, this follows from Euclid's lemma: $$p\mid ab\implies p\mid a\lor p\mid b$$.

That's because $$10=2\cdot5$$. So just apply the lemma to $$2$$ and $$5$$, to get $$2\mid n\land 5\mid n$$.

From this it follows easily that $$10\mid n$$, since the prime factorization of $$n$$ must contain a $$2$$ and a $$5$$.

"I think the fact that 10 is a product of unique primes"

Well, all numbers are products of unique primes.

But that it is a product of single powers of prime is key. That is: $$10 = 2*5$$ and none of the primes have higher power than $$1$$.

$$10|n^4$$ means $$2|n^4$$ means $$2|n$$ and the same thing for $$5$$ ($$5|n^4$$ so $$5|n$$) so $$2*5 =10|n$$.

But this wouldn't hold true for a prime factor with a power more than $$1$$.

For example, if $$20|n^4$$ and $$20 = 2^2* 5$$, then $$2|n^4$$ and $$2|n$$, but $$2^2|n^4$$ does not mean $$2^2|n$$. For example, if $$n = 10$$ then $$20|10^4=10,000,$$ but $$20\not \mid 10$$.

However we can prove if $$20|n^4$$ then $$10|n$$.

In general, if $$\prod p_i^{k_i} = n^m$$ then $$\prod p_i |n$$.