Question 1209939
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If m is a positive integer and sqrt(4m^2+29) is an integer, then what is m?
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<pre>
If   {{{sqrt(4m^2+29)}}}  is an integer number n, then

    4m^2 + 29 = n^2

    29 = n^2 - 4m^2

    29 = (n-2m)*(n+2m).


29 is a prime number, so it has two possible decompositions into the product of prime factors 

        29 = 1* 29   or  29 = (-1)*(-29).


Therefore, we have 4 systems of linear equations to analyze

    n - 2m =  1,    (1)
    n + 2m = 29,    (2)

    n - 2m =  -1,   (3)
    n + 2m = -29,   (4)

    n - 2m =  29,   (5)
    n + 2m =   1,   (6)

    n - 2m = -29,   (7)
    n + 2m =  -1.   (8)



From system (1), (2), by subtracting equations, we have 

    2m - (-2m) = 29-1,  4m = 28,  m = 28/4 = 7  and then  n = 1 + 2m = 1 + 2*7 = 15.

So, the solution pair is (m,n) = (7,15), and it works properly: 

    {{{sqrt(4*7^2+29)}}} = {{{sqrt(4*49+29)}}} = {{{sqrt(225)}}} = 15.



From system (3), (4), by subtracting equations, we have 

    2m - (-2m) = -29-(-1),  4m = -28,  m = -28/4 = -7.  

It does not work, since the number m should be positive, by the condition.



From system (5), (6), by subtracting equations, we have 

    2m - (-2m) = 1-29,  4m = -28,  m = -28/4 = -7.  

It does not work, since the number m should be positive, by the condition.



From system (7), (8), by subtracting equations, we have 

    2m - (-2m) = -1 - (-29),  4m = 28,  m = 28/4 = 7  and then  n = 1 + 2m = 1 + 2*7 = 15.

So, the solution pair is (m,n) = (7,15), the same as we got from system (1), (2), 
and it works properly:  {{{sqrt(4*7^2+29)}}} = {{{sqrt(4*49+29)}}} = {{{sqrt(225)}}} = 15.


<U>ANSWER</U>.  For the given problem, there is a unique solution for m.  It is m = 7.
</pre>

Solved.


I placed this my solution here after the solution by tutor @math_tutor2020 to make the analysis complete.



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My comment/response to the Edwin's comment, regarding my post.


<pre>
    For this given problem, considering all four decompositions

        29 = 1*29 = 29*1 = (-1)*(-29) = (-29)*(-1)


    is NECESSARY for the completeness of the analysis.


    So, all 4 (four) cases/decompositions MUST be considered, exactly as it was made in my post.
</pre>