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Given that , $ m^{2} + 105 = n^{2}$

$ \Rightarrow n^{2} – m^{2} = 105 $

$ \Rightarrow (n-m) (n+m) = 105 \quad \longrightarrow (1) $

We can factorize $105 = 1 \cdot  105 = 3 \cdot 35 = 5 \cdot 21 = 7 \cdot 15 = 15 \cdot 7  = 21 \cdot 5 = 35 \cdot 3 = 105 \cdot 1$

Now, from equation $(1),$ we get

  • When $n-m = 1, n+m=105 \Rightarrow n=53, m=52 $
  • When $n-m=3, n+m=35 \Rightarrow n=19, m=16 $
  • When $n-m=5, n+m=21 \Rightarrow n=13, m=8 $
  • When $n-m=7, n+m=15 \Rightarrow n=11, m=4 $
  • When $n-m=15, n+m=7 \Rightarrow n=11, m=-4 $
  • When $n-m= 21, n+m=5 \Rightarrow n=13, m=-8 $
  • When $n-m=35, n+m=3 \Rightarrow n=19,m=-16 $
  • When $n-m=105, n+m=1 \Rightarrow n=53, m=-52 $

Since$,m$ and $n$ are positive integers. So, only $4$ pairs are possible.

Correct Answer $:4$

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