#(a-b)(a-c)(a-d)(b-c)(b-d)(c-d) -= 0 (mod 3)# Can anybody help me with this discrete mathematics question?

Question

Given that #a,b,c,d in ZZ# , show that
#(a-b)(a-c)(a-d)(b-c)(b-d)(c-d) -= 0 (mod 3)#

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Answer 1 · 2018-05-02T21:16:00

See explanation...

What are the values of #a, b, c, d# modulo #3# ?

They can only take values congruent to #0#, #1# or #2#.

So by the pigeonhole principle, at least two of #a, b, c, d# are congruent modulo #3#.

Then their difference is a multiple of #3# and hence:

#(a-b)(a-c)(a-d)(b-c)(b-d)(c-d) -= 0# (mod #3#)