# CAT 2019 Quant Question with Solution 38

Question:
How many pairs $(\mathrm{m}, \mathrm{n})$ of positive integers satisfy the equation $\mathrm{m}^{2}+105=\mathrm{n}^{2}$ ?

Shortcut:

Number of pairs = $\frac{number\ of\ factors\ 105}{2}$

$105=3\times 5\times 7$

Number of factors = $2\times 2\times 2=8$

Hence, required number of pairs =8/2 =4

Detailed Explanation:

${{\text{m}}^{2}}+105={{\text{n}}^{2}}$

$\Rightarrow {{n}^{2}}-{{m}^{2}}=105$

$\Rightarrow \left( n-m \right)\left( n+m \right)=105$

Since m and n are positive integers, $\left( n-m \right)<\left( n+m \right)$

Splitting 105 in two factors, we get

$\Rightarrow \left( n-m \right)\left( n+m \right)=1\times 105$

For $\left( n-m \right)=1$ and $\left( n+m \right)=105$, $(m,n)=(52,53)$

$\Rightarrow \left( n-m \right)\left( n+m \right)=3\times 35$

For $\left( n-m \right)=3$ and $\left( n+m \right)=35$, $(m,n)=(16,19)$

$\Rightarrow \left( n-m \right)\left( n+m \right)=5\times 21$

For $\left( n-m \right)=5$ and $\left( n+m \right)=21$, $(m,n)=(8,13)$

$\Rightarrow \left( n-m \right)\left( n+m \right)=7\times 21$

For $\left( n-m \right)=7$ and $\left( n+m \right)=21$, $(m,n)=(4,11)$

Hence there are four pairs.

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