GB-41-05-001 · Algebraic divisibility proof

For any integer n, prove that $(2n+9)^2-(2n+3)^2$ is divisible by 6.

Algebraic Proof (Prove a Multiple) Booklet p. 32–33

Method

1) Form expression

2) Expand using grids

3) Simplify

4) Factorise out the divisor

5) Write a conclusion

Worked Example (GB-41-05-001)

Form the expression

(2n+9)^2-(2n+3)^2

. Expand each square using a grid:

×	$2n$	$9$
$2n$	$4n^2$	$18n$
$9$	$18n$	$81$

×	$2n$	$3$
$2n$	$4n^2$	$6n$
$3$	$6n$	$9$

Subtract the whole second expansion:

(4n^2+36n+81)-(4n^2+12n+9)=24n+72

Factorise out the required divisor:

24n+72=6(4n+12)

For integer n,

4n+12

is an integer. The expression is therefore divisible by

6

Ans =

12(2n+6)