Let a and b be two positive integers, where a ≥ b. Find all pairs a, b such that their sum, their positive difference, their product, and their quotient add to 36

Answer 1

a+b + a-b + ab + a/b = 36
a(2 + b + 1/b) = 36
So, a is a factor of 36
Also, note that b + 2 + 1/b = (√b + 1/√b)^2
Looks like (2 + b + 1/b) must be a perfect square, so 1,4,9
I guess the only pair I see is a=9, b=1

Answer 2

Thanks

Answer 3

interpretation:

(a+b) + (a-b) + (ab) + (a/b) = 36
2a + ab + a/b = 36
2ab + ab^2 + a = 36b
ab^2 + a = 36b - 2ab
a(b^2 + 1) = 2b(18 - a)
a/(18-a) = 2b/(b^2 + 1)

I tried b = 1, then a = 9
b = 2, then a = 8
b = 5, then a = 5
could not find any others.

so (9,1), (8,2), (5,5)

testing: e.g. (8,2)
8+2 + 8-2 + (8)(2) + 8/2
= 16 + 16 + 4
= 36

Answer 4

Nice save