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$\forall$ $a$, $n$ $\in \mathbb Z$, if $a|n^2$ and $a\le n$, then $a|n$.

I've tried:-

From $ ∣

^ 2$

: Since $ ∣ ^2$, we can write:

$n ^ 2 =ak$, for some integer $$.

Assuming $ =

$

: For integers $$ and $$ where $0 ≤ < $ . This means we can express $$ in terms of $$:

$ =

$

Squaring $$ :

$$^2=(

) ^2=

^ 2 ^ 2

2

^ 2$$

$$\implies a^2+2aqr+r^2=ak$$

$$\implies a^2+2aqr+r^2-ak=0$$ $\therefore r^2\text{ mod }a=0$

I've to prove that $r\text{ mod }a=0$, and I'm back to where I started.

I've tried to prove this with the unique factorization theorem as well, but I can't find the right statements at all. The assertion seems simple, but I can't prove it at all.

Integreek
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Vince
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    Though $(-4)\mid 2^2$, $-4 \le 2$, and $(-4) \not\mid 2$. – peterwhy Sep 21 '24 at 04:12
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    For just positive $a$ and $n$, consider $a=4=2^2$ and $n=6=2\times 3$. – John Omielan Sep 21 '24 at 04:20
  • The positive counterexample comes from your attempt, by picking $r=2$, $a=4$ to satisfy both $r^2\bmod a = 0$ and $r\bmod a \ne 0$, with $n = a1+r = 6$. – peterwhy Sep 21 '24 at 04:24
  • I think $a$ should be a prime number for the statement to hold. If that is indeed the case, then you can prove it by writing the prime factorization of $n$. – Integreek Sep 21 '24 at 04:32
  • If we assume $a$ is a prime, OP's proof works. The key argument at the end would be that $r^2 \equiv 0 \pmod a$ implies $r =0$ since $\mathbb{Z}_{a}^{\times}$ is a field. If $a$ is not a prime, as others already have pointed out, the result is simply false. – huh Sep 21 '24 at 04:39
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    @MathGuy Note that if $a$ is required to be a prime number, then the $a\le n$ condition is not required. Also, we can basically just use Euclid's lemma to prove that $a\mid n$ rather than the prime factorization of $n$. – John Omielan Sep 21 '24 at 04:44
  • You used the following reduction $,a\mid nk\iff a\mid (n\bmod a)k,$ to reduce to the case $,n< a,,$ where, next, we can similarly reduce via $:!\ a\mid nk\iff a\mid (a\bmod n)k.,$ Now, if $(a,n)=1,,$ the last reduction will continually reduce $n$ to a nonzero value, so will eventually reach $,n=1,,$ which yields a proof of Euclid's Lemma $,(a,n)=1,, a\mid nk\Rightarrow a\mid k.,$ See the linked dupe for further details, including a constructive interpretation (Gauss's Lemma). $\ \ $ – Bill Dubuque Sep 21 '24 at 05:12
  • The prior method fails if $(a,n)>1$ in which case there are easy counterexamples. $\ \ $ – Bill Dubuque Sep 21 '24 at 05:31
  • Generally $\ a\mid nk\iff a\mid (a,n)k\iff a/(a,n)\mid k\ \ $ – Bill Dubuque Sep 21 '24 at 05:37
  • @peterwhy, how do you find counterexamples so easily? I can't find the right one at all and get stuck at proofs. – Vince Sep 21 '24 at 08:34
  • @huh what do you mean by:-r^2≡0(mod a) implies r=0 since Z×a is a field – Vince Sep 21 '24 at 08:36
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    @Vince don't worry, such observations will strike you as you practise more. As a general strategy, if proving a statement to be true seems difficult, then try to search for a counter example. – Integreek Sep 21 '24 at 10:07
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    @Vince, I referred to $\mathbb{Z}_a^{\times}$ as a field for convenience, but you don't need to consider the full assertion. The point is that when $a$ is a prime number, the residue system ${1, \ldots, a-1}$ contains no zero divisors. This means that if you have an equation of the form $x y \equiv 0 \pmod{a}$, then either $x \equiv 0 \pmod{a}$ or $y \equiv 0 \pmod{a}$. It is not difficult to prove this and you can apply it directly to your question (again, only when $a$ is a prime). – huh Sep 21 '24 at 15:03

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