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# help

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

May 28, 2020

#1
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Hint: (𝑚−𝑛)(𝑚+𝑛)=3⋅5⋅7
May 28, 2020
#2
+1

I believe that should be (n + m)(n – m) to keep the polarity straight... i.e., (n – m), not (m – n) because

the problem says m2 + 105 = n2        You would subtract m2 from both sides and get 105 = n2 – m2

.

Guest May 28, 2020
#3
+21953
+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:  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.

May 28, 2020
#4
+25541
+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

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

May 28, 2020