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Question 1162885: For how many integers n is n^2 + 18n + 13 a perfect square?
Found 3 solutions by solver91311, jim_thompson5910, ikleyn:
Answer by solver91311(24713)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
This is something you do through trial and error. Though there may be a clever proof or trick to be able to find all values of n quickly (I'm blanking on what that method would be).
I used a perl script to try out values of n from -1000 to 1000. The only two integers that make n^2 + 18n + 13 to be a perfect square are n = -27 and n = 9
If n = -27,
n^2 + 18n + 13 = (-27)^2 + 18(-27) + 13 = 256
If n = 9,
n^2 + 18n + 13 = (9)^2 + 18(9) + 13 = 256
Both values of n lead to the same perfect square.
It's quite possible that there may be more solutions. I'm not entirely sure as I only tested from n = -1000 to n = 1000.
Answer by ikleyn(52814)  (Show Source):
You can put this solution on YOUR website! .
It looks AMAZINGLY, but there is the way to solve the problem formally, in strict Algebra logic, without trials and errors.
Let n^2 + 18n + 13 = m^2 be a perfect square.
Then
(n + 9)^2 - 68 = m^2
(n + 9)^2 - m^2 = 68
(n + m + 9)*(n - m + 9) = 68.
Decompositions for 68 are 1*68, 2*34, 4*17, 17*4, 34*2 and 68*1.
For each decomposition, we have the system of equations
n + m + 9 = 1
n - m + 9 = 68
n + m + 9 = 2
n - m + 9 = 34
n + m + 9 = 4
n - m + 9 = 17
n + m + 9 = 17
n - m + 9 = 4
n + m + 9 = 34
n - m + 9 = 2
n + m + 9 = 68
n - m + 9 = 1
Easy analysis shows that some of these systems produce non-integer solution.
The only system, which produces appropriate integer solution, is THIS
n + m + 9 = 34
n - m + 9 = 2
The solution is n = 9, m = 16.
Therefore, n = 9 is the solution to the problem.
If you analyze the similar systems with decomposition of the number 68 into the product of negative factors,
you will find another solution n = -27.
In all, there are 2 (two) such numbers n, 9 and -27.
Solved.
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