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$a^{2} - 1 equiv 0pmod{mcdot a}$ same as $a^{2} - 1 equiv 0 pmod{m}$?

Given that $a$ and $m$ are relatively prime, is it possible for the above equation to hold true? My rational behind this is the following:

Because $a^{2} - 1 equiv 0 pmod{mcdot a}$, this implies that $mcdot amid left(a^{2} - 1right)$, which implies $a^{2} - 1 = lcdot mcdot a, l in mathbb Z$

Is it wrong to then say that: let $t = lcdot a, t in mathbb Z$. Hence, we have $a^{2} - 1 = tcdot m$, thus implying that $a^{2} - 1 equiv 0 pmod {m}$

Note that $a$ is a positive integer in this case

EDIT: What I meant to ask was does $a^{2} - 1 equiv 0 pmod{mcdot a}$ imply $a^{2} - 1 equiv 0 pmod{m}$?

It is not the same: it only goes in one direction. Yes, it is possible they are both true, but not a universal certainty. And, I don't think it has much to do with $a^2-1$.

If $x equiv 0 bmod ma$ then $ma mid x$. Since clearly $m mid ma$ we have $m mid x$ and so $x equiv 0 bmod m$. For your case, let $x=a^2-1$, but this really doesn't matter: it's still true.

The other direction may be false. For instance $10 equiv 0 bmod 5$ but $10$ is not congruent to $0 bmod 15$. Or, take my example from my comment with $a=4$ and $m=3$: $15equiv 0 bmod 3$ yet $15$ is not congruent to $0 bmod 12$.

Edit: I've interpreted your question "are they the same" as asking "is there an if and only if in between."

This statement is not biconditional.

To be precise,

if $a^2-1equiv 0mod (m*a)$, then $a^2-1equiv 0mod a$ .

if $a^2-1equiv 0mod (a)$, then it is not always true that $a^2-1equiv 0mod (m*a)$.

Your logic is correct, however some constraints may be removed. More generally, $m$ and $a$ do not have to be coprime,it does not have to be a, and the exponent can be any integer $n$, not just $2$.

A more general statement is as follows:

If $a^n-1equiv 0mod N$, then $a^n-1equiv 0mod f$, for every factor $f$ of $N$.

Proof. We can see that $N|a^n-1$. Therefore all of its factors must too. $f|a^n-1Longleftrightarrow a^n-1equiv 0mod f$.

I suggest looking into the concept of multiplicative order for a more thorough understanding.

Of course $,mamid a^2-1 ,Rightarrow, mmid a^2-1,$ is always true, $ $ by $ (ma)n = m(an)$.

The converse is true $iff m = 0.,$ For if $,m=0,$ both become sides are identical: $,0mid a^2-1$.

And if $,mneq 0,$ then the RHS: $,mmid a^2-1,$ holds for $,a = 1+km equiv 1pmod{!m}.,$

But the LHS $,Rightarrow, amid a^2-1,Rightarrow, amid 1,,$ falsified by choosing $,k,$ so $,a = 1+km > 1.$

$a^2 -1 pmod {ma} implies ma|a^2 - 1 implies m|a^2-1$ so that one direction always holds if it ever holds.

However $ma|a^2 - 1 implies exists k: a^2 - 1 = mak$ so $ a -frac 1a =mkin mathbb Z$ which implies $a equiv pm 1$ so $a^2 - 1 pmod {ma}$ only if $a = pm 1$.

So, yes, IF $a^2 -1 pmod {ma}$ then $a^2 -1 pmod m$. BUT $a^2 - 1 pmod{ma}$ only when $a = pm 1$.

The other way:

$a^2 -1 equiv pmod m implies a^2 - 1equiv km pmod {am}$ for some integer $k: 0le k< m$ (and chinese remainder theorem says that such a $k$ is unique) but there is no reason to assume $k =0$ (and lots of reason to assume it isn't).

So so any $a^2 -1 pmod {m}$ where $anot equiv pm 1 pmod m$ will be a counter example.

Fo instance $5^2 -1 equiv 0 pmod {24}$ and $gcd(a,5) = 1$ but $5^2 -1 not equiv 0 pmod {5*24 = 120}$