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Let $p$ be any prime. Show that the remainder when $(p-1)!$ is divided by $p(p-1)$ is $p-1$

I have attached my solution below. Can someone please confirm it, and if you find any mistake in it, please do tell.

Thanks in advance

From wilson's theorem, we know :

$$(p-1)! ≡ -1 ≡ p-1 (mod\ p)$$

$$(p-2)! ≡ 1 (mod\ p)$$

so, we get,

$$ (p-2)! = pk +1 $$ $$ (p-1)! = p(p-1)k + (p-1)$$

and clearly, the remainder is $p-1$,

Hence proved

Bill Dubuque
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Pranav Choudhary
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  • Please identify which step(s) in the proof you have doubts about, and why. If there are no such doubts then there is no need to post the question. – Bill Dubuque Apr 19 '21 at 18:35
  • Perfect................. – DanielWainfleet Apr 19 '21 at 18:49
  • It is correct. You are essentially applying CRT to solve $x\equiv -1\pmod{p},\ x\equiv 0\pmod{p-1},,$ which is easy since one modulus is $\equiv \pm1 $ mod the other, so in the CRT formula we only need to invert $\pm 1$, see Easy Inverse CRT – Bill Dubuque Apr 19 '21 at 19:04
  • You should justify why cancelling $,p-1,$ from the first congruence yields an equivalent congruence (hint: this can fail if the fatcor being cancelled is not coprime to the modulus), and you should explicitly signify such equivalences between sll congruences (use $!\iff!$ to relate them or use natural language) else the proof may be insufficient – Bill Dubuque Apr 19 '21 at 19:05

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