<s>[INST] Provide a counterexample: If $n^2-1$ is divisible by $5$, then $n$ is divisible by $2$ or $3$ Provide a counterexample: If $n^2-1$ is divisible by $5$, then $n$ is divisible by $2$ or $3$
My book doesn't have an answer to this question, but I think it's $n=6$. 
Since $6^2-1=35$, which is divisible by $5$ but not by $2$ or $3$. Is this the right way to solve these counterexample problems? Are these problems solved by just plugging in numbers until you get to the right one?
 [/INST] $n^2 - 1= (n-1)(n+1)$ so if $5|n^2 -1$, as $5$ is prime we know either $5|n-1$ or $5|n+1$.
Now if we're clever we can try to take counter examples $n=6$ or $4$.  nope.  $n = 11$ or $9$.  Oh, wait there's a counter example. $n=11$.  (Also $n=1$ and $5|0$ is a counter example.)
But plugging in numbers in the hope of a counterexample is ineffiecient.  What if the first counter example is really huge? Or what if the statement is true and we never find a counter example?
we should evaluate further.
So $n\equiv \pm 1 \pmod 5$.  And we are asked to show or find a counter example that $n \equiv 0 \pmod 3$ or $n \equiv 0\pmod 2$.
Now By Chinese remainder theorem $\gcd(5,2); \gcd(5,3); \gcd(2,3)=1$ so ther is a unique solution to 
$n \equiv a \pmod 5$ and $n \equiv b  \pmod 3$ and $n \equiv c \pmod 2$ will have a unique solution $\mod 30$ and we just need to choose $a = \pm 1$ and $b \ne 0$ and $c \ne 0$.
Example $n \equiv 1 \pmod 5$ and $n\equiv 1 \pmod 3$ and $n\equiv 1 \mod 2$ will have solutions and that solution would be a counter example.
We don't even have to find it!  We know it exists!
.... but it's $1 \pmod {30}$ so $n =1$ is counter example.  $n = 31$ is a counterexample and $n = 30k + 1$ is a counter example.  $(30k + 1)^2- 1 = 900k^2 + 60k + 1 - 1= 5(180k^2 + 12k)$ but $2|30k$ and $3|30k$ so $2,3\not \mid 30k + 1$).
We can find other counter examples.
$n \equiv -1 \pmod 5$ and $n \equiv 1 \pmod 3$ $n\equiv 1 \pmod 2$ would give us $n \equiv 19\pmod 30$ so $19, 49, 79$ etc are counter examples.
And $n \equiv 1 \pmod 5$ and $n \equiv -1 \pmod 3$ and $n \equiv 1 \pmod 2$ will have solution $n \equiv 11 \pmod 30$.  So $11, 41, 71$ etc are counter examples.
And $n \equiv -1\pmod 5$ and $n \equiv -1 \pmod 3$ and $n \equiv 1 \pmod 2$ will have soultion $n \equiv 29 \pmod 30$.  So $29, 59, 89$ etc are counter examples.
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