#3**+1 **

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: m^{2} + 105 = n^{2}

m^{2} + 105 = (m + 3)^{2}

m^{2} + 105 = m^{2} + 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.

geno3141 May 28, 2020

#4**+2 **

**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**

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

heureka May 28, 2020