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Is The Difference Between Two Square Numbers Always Odd?

Published Aug 29, 2025 5 min read
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No, the difference between two square numbers is not always odd . The result depends entirely on the parity (whether a number is even or odd) of the two integers that are being squared.

Case Parity of First Number Parity of Second Number Example Calculation Resulting Parity
Case 1 Odd (oo π‘œ ) Odd (oo π‘œ ) 52βˆ’32=25βˆ’9=165 squared minus 3 squared equals 25 minus 9 equals 16 52βˆ’32=25βˆ’9=16 Even
Case 2 Even (ee 𝑒 ) Even (ee 𝑒 ) 42βˆ’22=16βˆ’4=124 squared minus 2 squared equals 16 minus 4 equals 12 42βˆ’22=16βˆ’4=12 Even
Case 3 Odd (oo π‘œ ) Even (ee 𝑒 ) 52βˆ’42=25βˆ’16=95 squared minus 4 squared equals 25 minus 16 equals 9 52βˆ’42=25βˆ’16=9 Odd
Case 4 Even (ee 𝑒 ) Odd (oo π‘œ ) 42βˆ’32=16βˆ’9=74 squared minus 3 squared equals 16 minus 9 equals 7 42βˆ’32=16βˆ’9=7 Odd

The mathematical proof

We can use fundamental principles of number theory to prove these outcomes. First, recall the following properties of even and odd numbers:

  • An even number can be expressed as 2k2 k

    2π‘˜

    for some integer kk

    π‘˜

    .

  • An odd number can be expressed as 2k+12 k plus 1

    2π‘˜+1

    for some integer kk

    π‘˜

    .

  • The square of an even number is always even: (2k)2=4k2=2(2k2)open paren 2 k close paren squared equals 4 k squared equals 2 open paren 2 k squared close paren

    (2π‘˜)2=4π‘˜2=2(2π‘˜2)

    .

  • The square of an odd number is always odd: (2k+1)2=4k2+4k+1=2(2k2+2k)+1open paren 2 k plus 1 close paren squared equals 4 k squared plus 4 k plus 1 equals 2 open paren 2 k squared plus 2 k close paren plus 1

    (2π‘˜+1)2=4π‘˜2+4π‘˜+1=2(2π‘˜2+2π‘˜)+1

    .

We can also use the algebraic identity for the difference of squares:a2βˆ’b2=(aβˆ’b)(a+b)a squared minus b squared equals open paren a minus b close paren open paren a plus b close paren

π‘Ž2βˆ’π‘2=(π‘Žβˆ’π‘)(π‘Ž+𝑏)

This identity is key to understanding why the difference of squares is not always odd. The parity of the result depends on the parity of the factors (aβˆ’b)open paren a minus b close paren

(π‘Žβˆ’π‘)

and (a+b)open paren a plus b close paren

(π‘Ž+𝑏)

. The factors will always have the same parity (both even or both odd), which means their product will either be even (even Γ—cross

Γ—

even) or odd (odd Γ—cross

Γ—

odd).

Breaking down each case

Case 1: Both numbers are odd

Let the two odd numbers be represented by 2k+12 k plus 1

2π‘˜+1

and 2j+12 j plus 1

2𝑗+1

.The difference of their squares is:(2k+1)2βˆ’(2j+1)2open paren 2 k plus 1 close paren squared minus open paren 2 j plus 1 close paren squared

(2π‘˜+1)2βˆ’(2𝑗+1)2

Using the identity a2βˆ’b2=(aβˆ’b)(a+b)a squared minus b squared equals open paren a minus b close paren open paren a plus b close paren

π‘Ž2βˆ’π‘2=(π‘Žβˆ’π‘)(π‘Ž+𝑏)

, we get:((2k+1)βˆ’(2j+1))Γ—((2k+1)+(2j+1))open paren open paren 2 k plus 1 close paren minus open paren 2 j plus 1 close paren close paren cross open paren open paren 2 k plus 1 close paren plus open paren 2 j plus 1 close paren close paren

((2π‘˜+1)βˆ’(2𝑗+1))Γ—((2π‘˜+1)+(2𝑗+1))

=(2kβˆ’2j)Γ—(2k+2j+2)equals open paren 2 k minus 2 j close paren cross open paren 2 k plus 2 j plus 2 close paren

=(2π‘˜βˆ’2𝑗)Γ—(2π‘˜+2𝑗+2)

=2(kβˆ’j)Γ—2(k+j+1)equals 2 open paren k minus j close paren cross 2 open paren k plus j plus 1 close paren

=2(π‘˜βˆ’π‘—)Γ—2(π‘˜+𝑗+1)

=4(kβˆ’j)(k+j+1)equals 4 open paren k minus j close paren open paren k plus j plus 1 close paren

=4(π‘˜βˆ’π‘—)(π‘˜+𝑗+1)

Since the result is a multiple of 4 (and therefore a multiple of 2), the difference between two odd squares is always even.

Case 2: Both numbers are even

Let the two even numbers be represented by 2k2 k

2π‘˜

and 2j2 j

2𝑗

.The difference of their squares is:(2k)2βˆ’(2j)2open paren 2 k close paren squared minus open paren 2 j close paren squared

(2π‘˜)2βˆ’(2𝑗)2

=4k2βˆ’4j2equals 4 k squared minus 4 j squared

=4π‘˜2βˆ’4𝑗2

=4(k2βˆ’j2)equals 4 open paren k squared minus j squared close paren

=4(π‘˜2βˆ’π‘—2)

Since the result is a multiple of 4, the difference between two even squares is always even.

Case 3: One number is odd, the other is even

Let the odd number be 2k+12 k plus 1

2π‘˜+1

and the even number be 2j2 j

2𝑗

.The difference of their squares is:(2k+1)2βˆ’(2j)2open paren 2 k plus 1 close paren squared minus open paren 2 j close paren squared

(2π‘˜+1)2βˆ’(2𝑗)2

=(4k2+4k+1)βˆ’4j2equals open paren 4 k squared plus 4 k plus 1 close paren minus 4 j squared

=(4π‘˜2+4π‘˜+1)βˆ’4𝑗2

=4k2+4kβˆ’4j2+1equals 4 k squared plus 4 k minus 4 j squared plus 1

=4π‘˜2+4π‘˜βˆ’4𝑗2+1

=4(k2+kβˆ’j2)+1equals 4 open paren k squared plus k minus j squared close paren plus 1

=4(π‘˜2+π‘˜βˆ’π‘—2)+1

Since the result is an expression of the form 4(integer)+14 open paren integer close paren plus 1

4(integer)+1

, it is always odd.

Special case: The difference of consecutive squares

A specific scenario that often causes confusion is the difference of consecutive square numbers.If we take any two consecutive numbers, one must be even and the other must be odd.For example, consider the consecutive integers nn

𝑛

and n+1n plus 1

𝑛+1

.The difference of their squares is:(n+1)2βˆ’n2open paren n plus 1 close paren squared minus n squared

(𝑛+1)2βˆ’π‘›2

=(n2+2n+1)βˆ’n2equals open paren n squared plus 2 n plus 1 close paren minus n squared

=(𝑛2+2𝑛+1)βˆ’π‘›2

=2n+1equals 2 n plus 1

=2𝑛+1

Since 2n+12 n plus 1

2𝑛+1

is the definition of an odd number, the difference between any two consecutive perfect squares is always odd. This is likely where the misconception arises that the difference of any two square numbers must be odd.

Summary of findings

The parity of the difference between two square numbers depends on the parity of the original numbers being squared. It is not always odd.

  • The difference of two odd squares is even.
  • The difference of two even squares is even.
  • The difference of one odd and one even square is odd.
  • As a special case of the odd/even difference, the difference between two consecutive squares is always odd.
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