When is this congruence true for integer $a>1$?
$$(a-1)^{(a+1)} \equiv (a-1) \> mod \> (a^2-1)$$
My try:
$$(a-1)^{a} \equiv \> 1 \> mod \> (a+1)(a-1)$$
I do not know what to do now.
Thank you.
Noob question about an exponential congruence
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modular-arithmetic
modules
congruences
congruence-relations
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1It's not true for a = 3, because $2^4 \equiv 0 \not \equiv 2 \mod 8$. – 2017-01-29
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0Since $a-1$ divides the modulus, you have to divide that as well. So you end up with $(a-1)^a \equiv 1 \pmod{(a + 1)}$, which is false for $a=3$. – 2017-01-29
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0@quasi one can show that this is true exactly for $a = p - 1$, where $p$ is prime. – 2017-01-29
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0@Santiago -- I'm on my way out -- I'll look at this later. – 2017-01-29
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0Thank you! Then, do I have to use Fernat's Theorem? – 2017-01-29