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To further the hint of 3, 5, and 7.

Consider 3:

--  we know that n > m

--  let us say that n = m + 3

--  then, we have:  m² + 105  =  n²

                              m² + 105  =  (m + 3)²

                              m² + 105  =  m² + 6m + 9

                                       105  =  6m + 9

                                         96  =  6m

                                          16  =  m

--  if m = 16:  m = n + 3  =  19, so we have the solution:  m = 16, n = 19

Repeat the above steps for 5 and 7 to get the other two answers.

Find the solutions to \(m^2 + 105 = n^2\) in positive integers.

\(\begin{array}{|rcll|} \hline m^2 + 105 &=& n^2 \\ n^2-m^2 &=& 105 \\ \mathbf{(n-m)(n+m)} &=& \mathbf{105} \\ \mathbf{x*y} &=& \mathbf{105} \\ && \text{find all integer products }\ x*y = 105, \\ && \text{where}\ x=n-m,\ \text{ and }\ y=n+m \\ \hline \end{array} \)

All integer products

\(\begin{array}{|ccl|r|r|r|r|r|} \hline && x*y & x & y & n=\frac{y+x}{2} & m=\frac{y-x}{2} \\ \hline 105 &=& 1\times 105 & 1 & 105 & \frac{105+1}{2}=53 & \frac{105-1}{2}=52 & \mathbf{52^2+105 =53^2} \\ 105 &=& 3\times 35 & 3 & 35 & \frac{ 35+3}{2}=19 & \frac{ 35-3}{2}=16 & \mathbf{16^2+105 =19^2} \\ 105 &=& 5\times 21 & 5 & 21 & \frac{ 21+5}{2}=13 & \frac{ 21-5}{2}= 8 & \mathbf{ 8^2+105 =13^2} \\ 105 &=& 7\times 15 & 7 & 15 & \frac{ 15+7}{2}=11 & \frac{ 15-7}{2}= 4 & \mathbf{ 4^2+105 =11^2} \\ \hline \end{array}\)