Find the number of ordered pairs (a,b) of integers such that
\frac{a + 2}{a + 1} = \frac{b}{8}.
I don't know how to read that shorthand but I'm going to assume it means
(a + 2) b
––––––– = –––
(a + 1) 8
I don't know how to solve this, so, as my geometry teacher used to say
"Go as far as you can, then see how far you can go."
Let's try some numbers and see if we can find a pattern
When a is 0, we get 2 / 1 so in order to preserve equality b = 16
1 3 / 2 12
2 4 / 3 won't work, no integer is 1/3 of 8
3 5 / 4 10
4 6 / 5 won't work, no integer is 1/5 of 8
5 7 / 6 won't work
6 8 / 7 won't work
7 9 / 8 9
I think I can see intuitively that no a > 7 will result in an integer b.
Let's go the other direction from zero and see what happens.
When a is –1, we get 1 / 0 zero in denominator is a no-no
–2, we get 0 / –1 so b = 0
–3, we get –1 / –2 so 4
–4, we get –2 / –3 won't work
–5, we get –3 / –4 so 6
–6, we get –4 / –5 won't work
–7, we get –5 / –6 won't work
–8, we get –6 / –7 won't work
–9, we get –7 / –8 so 7
It looks like the equality stops working when (a+1)
is less than negative 8 and greater than positive 8.
I can't see a general pattern – other than, in the ones that work the denominator
is counting up in base 2 but I don't know if that means anything. If if this were my
homework, I'd go out on that limb and say that the answer is 8 ordered pairs.
.