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Conjecture: When $n \gt 2$ is an even number, $n\pm 1$ is a twin prime pair iff there exists a nontrivial solution of $X^{(n-2)n} = 1 \pmod {n^2 - 1}$.

Proof:

To see this suppose $n \pm 1 \in \Bbb{Z}$ are twin primes. Then by Euler's totient theorem $X^{n-2} = 1 \pmod {n-1}$ and $X^{n} = 1 \pmod {n+1}$. So that by the Chinese Remainder theorem those two equations multiply into $X^{(n-2)n} = 1 \pmod {n^2 -1}$.

Conversely, suppose that the even number $n \gt 2$ is such that a solution $X$ exists to the equation $X^{n(n-2)} = 1 \pmod {n^2-1}$.

Since $n$ is even $n \pm 1$ must be a coprime pair of integers: Fixing the coprime error. $d | n \pm 1 \implies n-1 = n+1 \pmod d \implies -1 = 1 \implies d = 2$ but $n$ is even, so $dx = n + 1 \implies 1 = 0 \pmod 2$. Done.

Since $n \pm 1$ are coprime now, we can use the reverse isomorphism from the Chinese Remainder Theorem and break into two equations: $X^{(n-2)n} = 1 \pmod {n\pm 1}$.

We feel the need to appeal to Euler's totient function. We know that the multiplicative group of the ring $\Bbb{Z}/(m)$ must have order $\phi(m)$ and that there is a formula for it: $\phi(m) = m \prod_{p |m}p^{e-1}(1 - 1/p)$ where $e$ in the formula is the max exponent of the prime $p$ dividing $m$.

Let $m = n^2 -1$. And $R = \Bbb{Z}/(m)$. Clearly $\phi(m) = |R^{\times}| \lt |R|$. Since in our smaller rings $X^{n(n-2)} = 1$, we have that

Apparently this conjecture is false. See below.

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    Why would the exponent being larger than something matter? – 2017-01-07
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    If $dx=n-1$ and $dy=n+1$, then $d(x-y)=-2$. You probably want to say that if $d\mid x$ and $d\mid y$, then $d\mid y-x$, so $d\mid 2$. As $d$ cannot be even, $d=1$. – 2017-01-07
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    @CurveEnthusiast you can deduce that $d\mid 2$ from $d(x-y) = -2$. – 2017-01-07
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    @Darth Geek, Exactly, that's the point (I just swapped $x$ and $y$). The question states that $d(x-y)=0$ instead. – 2017-01-07
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    Are you sure that $x^{n-2}\equiv 1\pmod{n^2-1}$ implies that $n-1$ and $n+1$ are both prime? – 2017-01-07
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    @ajotatxe I think OP means that $x^{n(n-2)} \equiv 1 \pmod{n^2-1}$ implies that $n\pm 1$ are primes. Whether or not that is true I cannot say. – 2017-01-07
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    No, why would it? – 2017-01-07
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    @RobArthan I see where the misunderstanding comes from. It comes from my bad writing. Making edits. – 2017-01-07
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    What exactly is the *question* that you are asking? – 2017-01-07

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Let $n=14$. Then $(n-2)n=12\cdot 14=168$, and $n^2-1=14^2-1=195$. Take $X=2$. Then $$2^{168}\equiv1\bmod{195}.$$ But $n-1=13$, while $n+1=15$, and 15 is clearly not prime. Your statement seems to be incorrect.

What I expect to be the correct statement:

For an even number $n$, the pair (n-1,n+1) is a twin prime pair if and only if $(\Bbb Z/(n^2-1)\Bbb Z)^\times$ has $(n-2)n$ elements.

In this case the first part of your proof works out. For the second part we can use that $\varphi(n^2-1)=(n-2)n$, from which follows that $\varphi(n-1)=n-2$ and $\varphi(n+1)=n$. This shows that they are prime.