--> With reference to the Wolstenholme's theorem -> I've tried the modulus thing but... no success..
--> Let a/b = 1 + 1/2 + 1/3 + 1/4 + ... + 1/(p-1) , where p≥5 is a prime...
--> Prove that :-> (i) p | a AND ALSO (ii) p² | a !
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Is it 9th class question ???
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Hey there,
If p >3 is congruent to 3 mod 4, there is an answer which involves only p(mod8)p(mod8) and h(mod4)h(mod4), where hh is the class number of Q(−p−−−√)Q(−p) . Namely one has (p−12)!≡1(modp)(p−12)!≡1(modp) if an only if either (i) p≡3(mod8)p≡3(mod8) and h≡1(mod4)h≡1(mod4) or (ii) p≡7(mod8)p≡7(mod8) and h≡3(mod4)h≡3(mod4).
The proof may not be original: since p≡3(mod4)p≡3(mod4), one has to determine the Legendre symbol
((p−12)!p)=∏x=1(p−1)/2(xp)=∏x=1(p−1)/2(((xp)−1)+1).((p−12)!p)=∏x=1(p−1)/2(xp)=∏x=1(p−1)/2(((xp)−1)+1). It is enough to know this modulo 4 since it is 1 or -1. By developing, one gets (p+1)/2+S(mod4)(p+1)/2+S(mod4), where S=∑x=1(p−1)/2(xp).S=∑x=1(p−1)/2(xp). By the class number formula, one has (2−(2/p))h=S(2−(2/p))h=S (I just looked up Borevich-Shafarevich, Number Theory), hence the result, since (2p)(2p) depends only on p(mod8)p(mod8).Hope this helps!
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