Hi! I've got a problem with this assignment. I really hope you can help me.
Find all such pairs of different natural numbers that
is divisible by
Thanks!
This looks like an interesting problem. If divides then it also divides and hence it divides . But , and therefore must divide .
But then . Thus , and we only have to investigate the possibilities a = 1, 2 or 3.
When a = 1, the condition that divides becomes: divides , so that must be a factor of 24. You can then check that there are only three such values for b (given that b is different from a). Similar calculations (I'll leave you to work out the details) show that there are no solutions for a=2, and only one solution for a=3.
Spoiler:
If 3b+13 divides 4(b–3) then it divides 12(b–3) = 4(3b+13) – 88, so it divides 88. The only divisors of 88 which are larger than 13 are 22, 44 and 88. The only one of those that gives a solution to the problem is 88 = 3x25 + 13, so b=25.
For a=2, you should get the condition that 2b+8 divides 4(b–2). Therefore b+4 divides 2(b–2) = 2(b+4) – 12. So b+4 divides 12. The only divisors of 12 which are greater than 4 are 6 and 12, and neither of these gives a solution to the problem.