Given:n∣mwherenandmareintegersgreaterthan1,and
a≡b(modm),whereaandbareintegers.
Toshow:a≡b(modn)
Proof:Sincen∣m,i.e,ndividesm,wehave−
m=nα,whereαisaninteger.................(A)
Alsoa≡b(modm)⟹mdivides(a−b)⟹m∣(a−b)
⟹a−b=mλ,whereλisaninteger.................(B)
Substitutem=nαinequation(B),weget−
a−b=(nα)λ
⟹a−b=n(αλ)(∵Multiplicationisassociative.)
⟹ndividesa−b
⟹n∣(a−b)
⟹a−b≡0(modn)
⟹a≡b(modn)
Hencetheresult.
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