- #1

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- TL;DR Summary
- Hi All, I would like to ask for help in proving the following congruence relation:

Let p be a prime number and let ##n## be a natural such that ##p < n < p^2##. Then

$$ {n-1 \choose p-1} {n \choose p-1} \equiv 0 (\mbox{mod p}) $$

Hi,

I would like to prove the following congruence relation:

Let p be a prime number and let ##n## be a natural such that ##p < n < p^2##.

Then

$$ {n-1 \choose p-1} {n \choose p-1} \equiv 0 (\mbox{mod p}) .$$

I am expecting it to have a rather trivial proof. Thanks in advance for any contribution, ...

Best Regards,

DaTario

I would like to prove the following congruence relation:

Let p be a prime number and let ##n## be a natural such that ##p < n < p^2##.

Then

$$ {n-1 \choose p-1} {n \choose p-1} \equiv 0 (\mbox{mod p}) .$$

I am expecting it to have a rather trivial proof. Thanks in advance for any contribution, ...

Best Regards,

DaTario