How many pairs of natural numbers are there such that the difference of whose squares is 63?

Correct Answer: Option 1 (3)

Step-by-Step Explanation:

Let the two natural numbers be $x$ and $y$, where $x>y$.

According to the given question, the difference of their squares is 63. We can set up the algebraic equation as:

$x^2-y^2=63$

Using the algebraic identity for the difference of two squares, $a^2-b^2=(a-b)(a+b)$, we can factorize the left-hand side:

$(x-y)(x+y)=63$

Since $x$ and $y$ are natural numbers ($x,y\in \mathbb{N}=\{1,2,3,...\}$), both factors $(x-y)$ and $(x+y)$ must be integers. Furthermore, since $x$ and $y$ are positive, $(x+y)>(x-y)>0$.

Now, let us find all pairs of positive integer factors $(a,b)$ such that $a\times b=63$ with $a<b$, where $a=x-y$ and $b=x+y$:

1. Case 1: $x-y=1$ and $x+y=63$

Adding the two equations: $2x=64\Rightarrow x=32$

Subtracting the two equations: $2y=62\Rightarrow y=31$

This gives the pair $(32,31)$.

2. Case 2: $x-y=3$ and $x+y=21$

Adding the two equations: $2x=24\Rightarrow x=12$

Subtracting the two equations: $2y=18\Rightarrow y=9$

This gives the pair $(12,9)$.

3. Case 3: $x-y=7$ and $x+y=9$

Adding the two equations: $2x=16\Rightarrow x=8$

Subtracting the two equations: $2y=2\Rightarrow y=1$

This gives the pair $(8,1)$.

Thus, there are exactly 3 such pairs of natural numbers: $(32,31)$, $(12,9)$, and $(8,1)$.