Problem 2

Find one pair of positive integers

a,b

such that

ab(a+b)

is not divisible by

7

, but

(a+b)^7-a^7-b^7

is divisible by

7^7

Step 1 of 3: Factor the seventh-power difference

Previous step Next step

In plain words

The seventh power supplies one factor of seven; the square supplies six more when its base is divisible by $7^3$ .

(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2

Detailed analysis

Expanding $(a+b)^7$ and collecting the middle terms gives the displayed identity. Since the first factor $7ab(a+b)$ contributes exactly one factor of $7$ when $7\nmid ab(a+b)$ , it remains to make the quadratic factor divisible by $7^3$ .