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Why is the Square of an Odd Number Always 1 More Than a Multiple of 4?

Published in Number Theory Property 3 mins read

The square of an odd number is always 1 more than a multiple of 4 because any odd number can be expressed algebraically, and when squared, the resulting expression consistently simplifies to a form that explicitly shows a term divisible by 4, with a remainder of 1.

This mathematical property is a fundamental concept often explored in number theory and algebra. Understanding it demonstrates how basic algebraic principles can reveal consistent patterns within number systems.

Understanding Odd Numbers Algebraically

To prove this property, we first need a general way to represent an odd number.

  • An even number can always be written as 2n, where n is any integer.
  • An odd number is simply one more or one less than an even number. Therefore, an odd number can be represented as 2n + 1 or 2n - 1.

For this proof, as seen in mathematical derivations (e.g., from MyTutor.co.uk), it is convenient to use the form (2n-1). This representation works for any integer value of n (where n ≥ 1 to ensure positive odd numbers, or n can be any integer for signed odd numbers).

The Algebraic Proof Unpacked

Let's take our general odd number, (2n-1), and square it:

  1. Start with the general form of an odd number:
    Let an odd number be represented by (2n-1), where n is an integer.

  2. Square the odd number:
    (2n-1)²

  3. Expand the expression:
    Using the algebraic identity (a-b)² = a² - 2ab + b²:
    (2n-1)² = (2n)² - 2(2n)(1) + (1)²
    = 4n² - 4n + 1

  4. Factor out 4 from the first two terms:
    Now, observe the terms 4n² - 4n. Both terms have a common factor of 4.
    4n² - 4n + 1 = 4(n² - n) + 1

Why 4(n²-n) is a Multiple of 4

The key to understanding why the square of an odd number is 1 more than a multiple of 4 lies in the term 4(n²-n).

  • Definition of a Multiple: A number is a multiple of 4 if it can be written as 4 * k, where k is an integer.
  • The term (n²-n): Since n is an integer, will also be an integer, and so will n²-n. Let k = n²-n. Since n is an integer, k will always be an integer.
  • Conclusion: Therefore, 4(n²-n) is explicitly 4 * k, where k is an integer. This means 4(n²-n) is always a multiple of 4.

With 4(n²-n) being a multiple of 4, and the +1 term remaining, the entire expression 4(n²-n) + 1 clearly shows that the square of any odd number is always 1 more than a multiple of 4.

Illustrative Examples

Let's test this with a few odd numbers:

Odd Number (2n-1) Value of n Square ((2n-1)²) Result (4(n²-n)+1) Is it 1 more than a multiple of 4?
1 n=1 1² = 1 4(1²-1)+1 = 4(0)+1 = 1 1 = 0 + 1 (0 is a multiple of 4)
3 n=2 3² = 9 4(2²-2)+1 = 4(4-2)+1 = 4(2)+1 = 8+1 = 9 9 = 8 + 1 (8 is a multiple of 4)
5 n=3 5² = 25 4(3²-3)+1 = 4(9-3)+1 = 4(6)+1 = 24+1 = 25 25 = 24 + 1 (24 is a multiple of 4)
7 n=4 7² = 49 4(4²-4)+1 = 4(16-4)+1 = 4(12)+1 = 48+1 = 49 49 = 48 + 1 (48 is a multiple of 4)

As shown in these examples, every time an odd number is squared, the result consistently follows the pattern of being 1 more than a number perfectly divisible by 4.

Conclusion

In summary, the property that the square of an odd number is always 1 more than a multiple of 4 is a direct consequence of its algebraic representation. By expressing any odd number as (2n-1) and expanding its square, we arrive at the form 4(n²-n)+1. The term 4(n²-n) is definitively a multiple of 4, leaving the +1 as the remainder, thus proving the statement.