A more efficient algorithm (of Gauss and later Kraitchik) would be to find solutions to the congruence x^2=y^2 (mod p) e.g.

38^2=3^2 (mod 1435)

It is trivial to then check if the factor (38+3 in this case) is a permutation of n/2 digits from the original number.

In reply to Factors by gumby
in thread Vampire Numbers by YuckFoo

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