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I managed to prove that $p^4$ divides ${p^2 \choose p}-p$ by writing ${p^2\choose p}-p=p{p^2-1\choose p-1}-p$ and then using gaussian pairing(multiplying the last and first terms in the numerator of ${p^2-1\choose p-1}$)to show the congruence holds true.

However I could not advance to $p^5$ divides ${p^2 \choose p}-p$. Any help would be appreciated.

Sahaj
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tensorman666
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  • You can add the attempts you tried – Sahaj Oct 15 '24 at 04:36
  • It would also help to know where the question comes from, to give some context to the kind of approach that might be useful (or to just see if the question has already been answered elsewhere). – ConMan Oct 15 '24 at 04:49
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    Special case $,k=2,,n=1,$ in this AoPS post on CAMO 2020 #2 which proves $$\large {p^k \choose {,pn}, }\equiv {\ !p^{k-1},\choose n,}\pmod{!p^{2k+1}}\qquad$$ See also here. $\ \ $ – Bill Dubuque Oct 15 '24 at 05:11
  • I think you had a typo in $\binom{p^2}{p} = p\binom{\color{red}{(p-1)^2}}{p-1}$. I think you meant $\color{red}{p^2-1}$. If that's not the case, please roll back the edit. – Sahaj Oct 15 '24 at 05:16

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