To solve the problem of finding the number of positive integers (x) such that (x^2 + 19x + 48) is a perfect square, we proceed as follows:
First, note that:
[x^2 + 19x + 48 = (x+3)(x+16)]
We need this product to be a perfect square. Let (a = x+3), then (x+16 = a + 13), so ((a)(a+13) = k^2) for some integer (k).
Rearranging ((a)(a+13)=k^2):
[4a^2 + 52a = 4k^2]
[ (2a + 13)^2 - 169 = 4k^2 ]
[ (2a + 13)^2 - (2k)^2 = 169 ]
[ (2a + 13 - 2k)(2a + 13 + 2k) = 169 ]
169 is (13^2), so its positive factor pairs are ((1,169)) and ((13,13)):
For ((1,169)):
[2a + 13 - 2k = 1]
[2a + 13 + 2k = 169]
Adding these equations: (4a + 26 = 170 \implies a = 36). Thus (x = a - 3 = 33).
For ((13,13)):
[2a + 13 - 2k = 13]
[2a + 13 + 2k = 13]
This gives (a = 0), which is invalid ((x=-3) not positive).
For (x>33), ((x+3)(x+16)) lies between ((x+9)^2) and ((x+10)^2) (consecutive squares), so it cannot be a square.
Thus, the only positive (x) is 33.
Answer: (\boxed{1})
(\boxed{1})
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