How many pairs of natural numbers are there such that the difference of their squares is 35?
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Understanding the Problem: Natural Number Pairs
The question asks us to find the number of pairs of natural numbers \((a, b)\) such that the difference of their squares is 35. Natural numbers are positive integers, i.e., {1, 2, 3, ...}. The condition given is:
\($a^2 - b^2 = 35$\)
Here, \(a\) and \(b\) must be natural numbers.
Factoring the Difference of Squares
We can factor the left side of the equation using the difference of squares identity, which states that \($x^2 - y^2 = (x-y)(x+y)$\). Applying this to our equation, we get:
\($(a-b)(a+b) = 35$\)
Analyzing the Factors of 35
Since \(a\) and \(b\) are natural numbers, \(a\) and \(b\) must be positive integers. Their sum \($a+b$\) must be a positive integer. Their difference \($a-b$\) must also be an integer. Since \((a-b)(a+b) = 35\), and 35 is positive, both factors \($a-b$\) and \($a+b$\) must have the same sign. Since \($a+b$\) must be positive (sum of natural numbers), \($a-b$\) must also be positive.
So, we are looking for two positive integer factors of 35 whose product is 35. The positive integer factors of 35 are 1, 5, 7, and 35.
Let \($x = a-b$\) and \($y = a+b$\). We have \($xy = 35$\), where \(x\) and \(y\) are positive integers. Also, since \(a\) and \(b\) are natural numbers, \($a+b > a-b$\) unless \(b=0\), but \(b\) must be a natural number (so \(b \ge 1\)). Thus, \($a+b > a-b$\) is true for natural numbers \(a, b\). So, we need to find pairs of factors \((x, y)\) of 35 such that \(xy=35\), \(x\) and \(y\) are positive integers, and \($x < y$\).
The possible pairs \((x, y)\) satisfying these conditions are:
- \((1, 35)\)
- \((5, 7)\)
Parity Check
Let's consider the sum and difference of these factors:
\($(a-b) + (a+b) = 2a$\)
\($(a+b) - (a-b) = 2b$\)
For \(a\) and \(b\) to be integers, both \($a-b$\) and \($a+b$\) must have the same parity (both even or both odd). If they have different parities, their sum and difference would be odd, leading to non-integer values for \(2a\) and \(2b\), and thus non-integer \(a\) and \(b\). The factors of 35 (1, 5, 7, 35) are all odd numbers. When we form pairs \((x, y)\) such that \(xy=35\), both \(x\) and \(y\) must be odd. Thus, both pairs \((1, 35)\) and \((5, 7)\) consist of two odd numbers, satisfying the parity requirement for \(a\) and \(b\) to be integers.
Solving for Natural Numbers a and b
Now we solve for \(a\) and \(b\) for each valid pair \((a-b, a+b)\):
Case 1: \($a-b = 1$\) and \($a+b = 35$\)
We have a system of linear equations:
- \($a - b = 1$\)
- \($a + b = 35$\)
Adding the two equations:
\($(a - b) + (a + b) = 1 + 35$\)
\($2a = 36$\)
\($a = 18$\)
Substitute \(a=18\) into the first equation:
\($18 - b = 1$\)
\($b = 18 - 1$\)
\($b = 17$\)
The pair is \((18, 17)\). Both 18 and 17 are natural numbers. This is a valid pair.
Case 2: \($a-b = 5$\) and \($a+b = 7$\)
We have a system of linear equations:
- \($a - b = 5$\)
- \($a + b = 7$\)
Adding the two equations:
\($(a - b) + (a + b) = 5 + 7$\)
\($2a = 12$\)
\($a = 6$\)
Substitute \(a=6\) into the first equation:
\($6 - b = 5$\)
\($b = 6 - 5$\)
\($b = 1$\)
The pair is \((6, 1)\). Both 6 and 1 are natural numbers. This is also a valid pair.
Counting the Valid Pairs
We found two pairs of natural numbers \((a, b)\) that satisfy the condition \($a^2 - b^2 = 35$\):
- \((18, 17)\)
- \((6, 1)\)
Therefore, there are 2 such pairs of natural numbers.
Revision Table: Key Concepts in Number Theory
| Concept | Description |
|---|---|
| Natural Numbers | The set of positive integers: {1, 2, 3, ...} |
| Difference of Squares | An algebraic identity: \($a^2 - b^2 = (a-b)(a+b)$\). Useful for factoring expressions involving squared terms. |
| Factors | Numbers that divide an integer evenly. For example, factors of 35 are 1, 5, 7, 35, -1, -5, -7, -35. |
| Parity | Whether an integer is even or odd. Used here to determine if \(a\) and \(b\) are integers based on the parity of \($a-b$\) and \($a+b$\). |
Additional Information: Related Number Problem Solving
Problems involving the difference of squares often require finding integer solutions to equations. Here are some related ideas:
- Integer Solutions: Sometimes problems ask for integer solutions instead of just natural numbers. This would include zero and negative integers. In our case, \($a-b$\) and \($a+b$\) could potentially be negative factors of 35 if we allowed integer \(a, b\), but the natural number constraint simplifies it.
- Simultaneous Equations: Solving for \(a\) and \(b\) from \($a-b=x$\) and \($a+b=y$\) is a standard technique for solving systems of linear equations. Adding and subtracting the equations are common methods.
- Prime Factorization: For larger numbers, finding factors systematically involves prime factorization. The prime factorization of 35 is \($5 \times 7$\).
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