Question 1162885
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            It looks  AMAZINGLY,  but there is the way to solve the problem formally,  in strict Algebra logic,  without trials and errors.


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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.
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Solved.