Algebraic proof with odd numbers

Algebraic proof

(1MA1) · Higher

ORIGINAL

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

Board and spec code confirmed against Edexcel 1MA1 · registry checked 2026-07-09How this checking works

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 $n$ .