HomeNumber Theory Proof The quadratic congruence x^2+1≡0 (mod p), where p is an odd prime, has a solution if and only if p≡1 (mod 4). August 24, 2021 0 Thequadraticcongruencex2+1≡0(modp),wherepisanoddprime,hasasolutionifandonlyifp≡1(mod4).Proof.Letabeanysolutionofx2+1≡0(modp),sothata2≡−1(modp).Becausepdoesnotdividea,theoutcomeofapplyingFermat′slittletheoremis1≡ap−1≡(a2)(p−1)/2≡(−1)(p−1)/2(modp)Therefore,mustbeoftheformthatis.Nowfortheoppositedirection.Intheproduct(p−1)!=1×2×…p−12×p+12×…(p−2)(p−1)Wehavethecongruencesp−1≡−1(modp)p−2≡−2(modp)p+12≡−p−12(modp)Rearrangingthefactorsproduces(p−1)!≡1×(−1)×2(−2)×…×p−12×(−p−12)(modp)≡(−1)p−12(1×2×…×p−12)2(modp)Becausethereare(p−1)/2minussignsinvolved.ItisatthispointthatWilson′stheoremcanbebroughttobear;for(p−1)!≡−1(modp),whence−1≡(p−1)!≡(−1)p−12[(p−12)!]2(modp)Ifweassumethatpisoftheform4k+1,then(−1)(p−1)/2)=1,leavinguswiththecongruence−1≡[(p−12)!]2(modp)Theconclusionisthattheintegersatisfiesthequadraticcongruencex2+1=0(modp).Letustakealookatanactualexample,say,thecasewhichisaprimeoftheform4k+1.Here,wehavep−12=6,anditiseasytoseethat6!=720≡5(mod13)And52+1=26≡0(mod13)Thus,theassertionthat[((p−1)/2)!]2+1=0(modp)iscorrectforp=13.See(Burton,2011,p.95) Tags: Number Theory Facebook Twitter
