How many ordered pairs (a; b) are there, with a and b positive integers greater than 1, such that a divides b + 1 and b divides a + 2 ? (a) 0 (b) 1 (c) 2 (d) 3 (e) 5 4.
Answers
Answer. The six ordered pairs are (1009, 2018),(2018, 1009),(1009 · 337, 674) =
(350143, 674),(1009 · 1346, 673) = (1358114, 673),(674, 1009 · 337) = (674, 350143),
and (673, 1009 · 1346) = (673, 1358114).
Solution. First rewrite the equation as 2 · 1009(a + b) = 3ab, and note that 1009
is prime, so at least one of a and b must be divisible by 1009. If both a and b are
divisible by 1009, say with a = 1009q, b = 1009r, then we have 2(q + r) = 3qr. But
qr ≥ q + r for integers q, r ≥ 2, so at least one of q, r is 1. This leads to the solutions
q = 1, r = 2 and r = 1, q = 2, corresponding to the ordered pairs (a, b) = (1009, 2018)
and (a, b) = (2018, 1009).
In the remaining case, just one of a and b is divisible by 1009, say a = 1009q. This
yields 2 · 1009(1009q + b) = 3 · 1009qb, which can be rewritten as
2 · 1009q = (3q − 2)b. Because the prime 1009 does not divide b, it must divide 3q − 2;
say 3q − 2 = 1009k. Then 1009k + 2 = 3 · 336k + k + 2 is divisible by 3, so
k ≡ 1 (mod 3). For k = 1, we get q = 337, a = 1009 · 337, b = 2q = 674. For k = 4,
we get q = 1346, a = 1009 · 1346, b = q/2 = 673. We now show there is no solution
with k > 4. Assuming there is one, the corresponding value of q is greater than 1346,
and so the corresponding
b =
2q
3q − 2
· 1009
is less than 673. Because b is an integer, it follows that b ≤ 672, which implies
1
b
≥
1
672
>
3
2018
, contradicting 1
a
+
1
b
=
3
2018
.
Finally, along with the two ordered pairs (a, b) for which a is divisible by 1009 and b
is not, we get two more ordered pairs by interchanging a and b.
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