Integer Problem

[Hockeyman]

Find all integers n>3 such that n-3 divides evenly into n^2 -n. The only one i can find that work are 4,5,6, and 9. I have tried all the way up to 20 and i have found no more solutions. Is there any other way to check?

 

[skeeter]

the danger of using guess and check to solve a problem such as this is that it sometimes leaves out a few other solutions.

first, do the long division ... divide n-3 into n^2 - n, you get n + 2 + 6/(n-3).

now, you need to find the values of n-3 that divide evenly into 6, so ...

all the integers that divide evenly into 6 are ... -1, 1, -2, 2, -3, 3, and -6, 6.

now set n-3 equal to each of the 8 integer divisors of 6 and solve for the 8 possible solutions.

 

[Hockeyman]

I don't entirely get how you did it, but our methods come out with the same answer. Thank you.