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Solution to Congruence

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Theorem
NumberTheory

Theorem

The congruence ax≔b(modm)ax\equiv b\pmod{m} where d=gcd(a,m)d=gcd(a,m) and d∣bd|b has a solution x≔a1āˆ’1b1(modm/d)x\equiv a_{1}^{-1}b_{1}\pmod{m/d}where a1=ad,b1=bda_{1}=\frac{a}{d},b_{1}=\frac{b}{d}.