Question:
How many pairs $(\mathrm{m}, \mathrm{n})$ of positive integers satisfy the equation $\mathrm{m}^{2}+105=\mathrm{n}^{2}$ ?
Correct Answer: 4
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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