Let $a$, $b$ $(b≤a)$ be two positive integers are not twin primes and $p$ is any prime number.
Is this congruence
$$ \binom{a^p}{b^p} \equiv \binom{a}{b}^p \pmod{p} $$
valid?
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4$\begingroup$ @MarkWildon how do you apply Lucas? $\endgroup$– Fedor PetrovCommented Aug 27, 2022 at 16:44
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$\begingroup$ I've deleted an inaccurate comment saying it followed from Lucas' Theorem. $\endgroup$– Mark WildonCommented Aug 27, 2022 at 18:00
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2 Answers
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No. For example, you may take $a=3$,$b=2$,$p=2$.
answered Aug 27, 2022 at 16:43
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If $a$ is equivalent to 3 mod 4 then $a\choose 2$ is odd and $a^2\choose 4$ is even. So the proposed congruence fails with $b=2$, $p=2$ and any such $a$.
answered Aug 27, 2022 at 17:21