Technique

Write an odd number as $2 m + 1$

Arithmetic • Number System • Divisibility

Statement

Every even integer can be written as

2 m

and every odd integer as

2 m + 1

for some integer

m

; substituting these forms turns a claim about all odd or all even numbers into an algebraic identity.

How to use it

1Use a different letter for each number: $x = 2 m + 1$ and $y = 2 n + 1$ , never $2 m + 1$ twice.
2Expand, then collect the multiples of the modulus you care about so that the remainder is left visible.
3To refute a universal claim one counterexample suffices; to establish one, the general form is required.

Show that the product of two consecutive integers is always even.

Step-by-step

1
Let the integers be $n$ and $n + 1$ .
Why: Consecutive integers have opposite parity, so a case split on $n$ is exhaustive.
2
If $n = 2 m$ , the product is $2 m (2 m + 1)$ ; if $n = 2 m + 1$ , it is $(2 m + 1) (2 m + 2) = 2 (2 m + 1) (m + 1)$ .
Why: Both cases carry an explicit factor of $2$ .
3
So the product is a multiple of $2$ in either case.

Answer

The product is always even

Watch out

An odd square is

1

more than a multiple of

8

, so a sum of two odd squares is

2

more than a multiple of

8

: even, but never a multiple of

4

. Reading the leading multiple of

4

in an expression of the form

4 k + 2

as making the whole thing divisible by

4

is the usual slip.

Also known as

parity, odd and even general form, 2m+1, parity argument, odd square modulo 8