Question 1209939
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Answer: <font color=red>m = 7</font>


Work Shown
k = some integer
k = sqrt(4m^2+29)
k^2 = 4m^2+29
k^2-4m^2 = 29
(k-2m)(k+2m) = 29
29 is prime so the only factors are 1 and 29.


k and m are integers, so k-2m and k+2m are also integers.
In order for (k-2m)(k+2m) = 29 to be the case, we'd need either
{{{system(k-2m = 1,k+2m=29)}}}
or
{{{system(k-2m = 29,k+2m=1)}}}
In either system of equations, adding straight down leads to 2k = 30 which solves to k = 15.


Then,
k-2m = 1 
15-2m = 1
-2m = 1-15
-2m = -14
m = -14/(-2)
<font color=red>m = 7</font>
Or,
k-2m = 29
15-2m = 29
-2m = 29-15
-2m = 14
m = 14/(-2)
m = -7
We ignore m = -7 since your teacher stated that m > 0
Check: sqrt(4m^2+29)=sqrt(4*<font color=red>7</font>^2+29)=15 is an integer.
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