Solution of the Week #473 - Always a Multiple

(a^7)*b-a*(b^7)

Let’s start by plugging a couple of numbers into the expression. When a,b is 2,1 it computes to 126, and when a,b is 3,1 it computes to 2184. These two numbers have a GCD of 42, so we have an upper bound, but will the expression ALWAYS come out to a multiple of 42 for any integers a,b? 42 is 2x3x7, so we can separately look at whether the expression is always a multiple of 2,3 and 7.

To help us we can factorise the expression thusly:

ab(a^6-b^6)

difference of squares

ab(a^3+b^3)(a^3-b^3)

sum/difference of cubes:

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

Unfortunately we can’t break it down any further than that.

Looking at whether the expression is always even is straightforward. If either a or b is even then the whole expression will be, but if they’re both odd, the term (a-b) will be even.

As to divisibility by 3, again if either a or b is, then the whole thing is. If a and b have the same remainder after dividing by 3, then (a-b) will be divisible by 3, and if they differ (1 and 2), then (a+b) will be divisible by 3.

Finally looking at divisibility by 7 is a bit more complicated.

We can make a table of all of the possibilities of remainders after dividing by 7 (the modulo), to see what we can determine:

As before if either a or b is 0, the whole expression is divisible by 7, and also if a=b or if a+b=7 (using the (a-b) and (a+b) factors).

Unfortunately for the remaining cases we need to use the messier factors. Lets assume b=1 and let a=2,3,4,5:

(a^2+a+1)=7,13,21,31 and (a^2-a+1)=1,7,13,21

So between the two factors one will always be divisible by 7 if b=1. That just places four more xs into the grid. However, since the expressions are symmetrical with respect to a and b, we can also put xs in the column where a=1. Also since the two factors differ only in the sign of the middle term, we can regard b=6 as b=-1, and a=6 as a=-1 and check off those cases too:

Finally allowing b=2 and a=3,4

(a^2+2a+4)=19,28 and (a^2-2a+4)=7,12

We can place xs in the remaining spots in row b=2, and by symmetry, the other six spots too.

So therefore (a^7)*b-a*(b^7) is always divisible by 2,3 and 7, and so the ultimate answer is 42.