Question 1162885
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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.
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