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Let $a \lt 105$ be positive integers, and such $\dfrac{a^2-1}{2014}\in \mathbb{Z}$

show that: $a=1$

since $2\mid a^2-1$, so $a$ is odd numbers,and $2014=2\cdot 19\cdot 53$

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  • $\begingroup$ Anything still need clearing up, network? $\endgroup$ – Gerry Myerson Oct 18 '15 at 11:33
  • $\begingroup$ Earth to network, come in please. $\endgroup$ – Gerry Myerson Oct 19 '15 at 12:24
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So you need $19\mid a^2-1$, which is $19\mid(a+1)(a-1)$, and since $19$ is prime, this implies $19\mid a+1$ or $19\mid a-1$; similarly, $53\mid a+1$ or $53\mid a-1$.

Now if you take the plus sign in both places, then $19\cdot53\mid a+1$, but $a$ has to be much bigger than $105$ for that to happen. Similarly if you take the minus sign in both places.

If you have $19\mid a+1$ and $53\mid a-1$, well, that second one says $a$ is $1$ or $54$, and neither satisfies $19\mid a+1$. If you have $19\mid a-1$ and $53\mid a+1$, the second implies $a=52$ or $a=105$, and neither satisfies $19\mid a-1$. Done.

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You should first solve $a^2=1 \pmod {53}$ and then make sure that all positive solutions $a < 105$ do not satisfy $a^2=1 \pmod {19}$ or $a^2=1 \pmod 2$

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