An alternative factorization

x1001i>0(x+ai)(mod101)\displaystyle { x }^{ 100 }-1\equiv \prod _{ i> 0 }^{ \quad }{ (x+{ a }_{ i }) } \pmod{101}

For positive integers ai,a_i, consider the congruence above. Observe that these are linear factors, so a standard factorization for integers will not work ((in fact you'll have a 50th50^\text{th} power in there).). Determine the minimum sum of all ai.{a}_{i}.

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