Algebraic proof with odd numbers

Algebraic proof

(1MA1) · Higher

ORIGINAL

Turn consecutive odd numbers into algebra, then expose the factor that proves divisibility.

Verified against Edexcel 1MA1 (2026 spec)

Question

Prove algebraically that the difference between the squares of any two consecutive odd integers is divisible by 8.

Every step worked, with the reasoning.

1
Let the consecutive odd integers be $2n + 1$ and $2n + 3$ , where $n$ is an integer.
Every odd integer has the form $2n+1$ ; adding 2 gives the next odd integer.
2
$(2n + 3)^2 - (2n + 1)^2$
Subtract the square of the first listed integer from the square of the second.
3
$= (4n^2 + 12n + 9) - (4n^2 + 4n + 1)$
Expand both brackets carefully.
4
$= 8n + 8 = 8(n + 1)$
Collect like terms and factor out 8.
5
Since $n + 1$ is an integer, $8(n + 1)$ is a multiple of 8; changing its sign if needed does not affect divisibility.
This also covers negative odd integers, where the non-negative difference is $8|n+1|$ .

Answer: The absolute difference is $8|n+1|$ , so it is divisible by 8 for every integer $n$ .

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