<s>[INST] Is it possible that $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$ same as $a^{2} - 1 \equiv 0 \pmod{m}$? $a^{2} - 1 \equiv 0\pmod{m\cdot a}$ 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 $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$,  this implies that $m\cdot a\mid \left(a^{2} - 1\right)$, which implies $a^{2} - 1 = l\cdot m\cdot a, l \in \mathbb Z$ 
Is it wrong to then say that: let $t = l\cdot a, t \in \mathbb Z$. Hence, we have $a^{2} - 1 = t\cdot m$, 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 $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$ imply $a^{2} - 1 \equiv 0 \pmod{m}$?
 [/INST] $a^2 -1 \pmod {ma} \implies ma|a^2 - 1 \implies m|a^2-1$ 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 =mk\in \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:
$a^2 -1 \equiv \pmod m \implies a^2 - 1\equiv km \pmod {am}$ for some integer $k: 0\le k< m$ (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 $a\not \equiv \pm 1 \pmod m$ will be a counter example.
Fo instance  $5^2 -1 \equiv 0 \pmod {24}$  and $\gcd(a,5) = 1$ but $5^2 -1 \not \equiv 0 \pmod {5*24 = 120}$ 
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