MathLabs

Problem 2

Find one pair of positive integers a,ba,b such that ab(a+b)ab(a+b) is not divisible by 77, but (a+b)7−a7−b7(a+b)^7-a^7-b^7 is divisible by 777^7.
Step 3 of 3: Verify the divisibility
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In plain words

The valuation count is transparent: one seven comes from the leading factor and six from the squared 737^3.

(a+b)7−a7−b7=7⋅342⋅(73)2=2⋅32⋅19⋅77(a+b)^7-a^7-b^7=7\cdot342\cdot(7^3)^2=2\cdot3^2\cdot19\cdot7^7
Detailed analysis

Substitution into the factorization gives a multiple of 71+6=777^{1+6}=7^7, and the previous step already showed that the first required product is prime to 77. Thus the pair solves the problem.