Question

if the sum of all positive rational numbers n such that
$\sqrt{{n}^{2} + 84n + 1941$
is an integer is
$\frac{189}{k}$
then find k?

Answer 1

Question seems some Typo :
But, I am giving answer as data provided in question.

Steps :
1) Let there be positive integer 'm' such that

$\sqrt{ {n}^{2} + 84n + 1941} = m \\ = > {n}^{2} + 84n + 1941 = {m}^{2}$

2) Now,

${n}^{2} + 84n + 1941 - {m}^{2} =0$
By observing this as Quadratic in 'n'.
Use Quadratic Formula to get ' n ' in terms of m
Then,

$n = \frac{ - 84 + \sqrt{ {84}^{2} - 4(1941 - {k}^{2}) } }{2} \: \:$

Since, n is positive, so we rejected negative part of solution.

$= > n = \frac{ - 84 + 2 \sqrt{ {m}^{2} - 177 } }{2} \\ \: \: \: \: \: \: \: \: \: n = - 42 + \sqrt{ {m}^{2} - 177 }$

3) Now,
Since, m is positive integer and n is positive rational number and looking at above expression .
=> n must be positive integer.

4) Hence, we have to find solutions such that 'n' and 'm' both are positive integers.
Let m^2-177= k^2 where m, k > 0 and integer.

Then, we have two cases possible.
m^2 - k^2 = 177
(m-k)(m + k) = 177 * 1
177 is prime.
=> m - k = 1 ----(1)
m + k = 177 -----(2)
Solution : m =( 177+1)/2 = 89 , k = 88

Case : 2
(m-k) (m + k) = 3 * 59
Now,
m -k = 3 ----(3)
m + k = 59 ----(4)
Solution : m = 31 ,k = 28

Here, if m = 31 then value of n becomes negative.
n = -42 + 28 = -14 ( rejected)
Extra :
m = 31 also gives integer solution to k, but here
n = -42 + 31 = -11 is negative and hence, it is not in our solution set.

5) Hence,
$n =( - 42 + \sqrt{ {89}^{2} - 177} ) \\ = > n \: = - 42 + \sqrt{7744} \\ = > n \: = - 42 + 88 = 46$

Therefore,
There exists only one solution such that 'n' and 'm' are positive integers .
That is, m = 89 , n = 46 .
There is only one positive rational value of n= 46 .

Answer 2

Answer:

46 is the answer I hope