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Prove if n is odd, then n^2 + n + 1 is odd

Step by step, with the reason for every step · Direct proof · Discrete Math

Step by step

The question

Question: prove that if n is odd, then n² + n + 1 is odd.

1 Write it with letters

n is odd, so n = 2a + 1, where a is a whole number.

n = 2a + 1

That is what odd means.

odd = 2 × a whole number + 1

Careful!

An example shows one number.

The letters show every number at once.

Careful: one example is not a proof

2 Put them in

Now put them in: n² + n + 1 becomes (2a + 1)² + (2a + 1) + 1.

n2 + n + 1 = (2a + 1)2 + (2a + 1) + 1

3 Work it out

(2a + 1)² means the bracket times itself.

= (2a + 1)(2a + 1) + (2a + 1) + 1

Every part times every part: 4a² + 2a + 2a + 1 + 2a + 1 + 1.

= 4a2 + 2a + 2a + 1 + 2a + 1 + 1

Like terms together: 4a² + 6a + 3.

= 4a2 + 6a + 3

4 Show the form

We take out 2 from all but the 1: 2(2a² + 3a + 1) + 1.

= 2(2a2 + 3a + 1) + 1

2a² + 3a + 1 is a whole number, because whole numbers added and multiplied stay whole.

2a² + 3a + 1 is a whole number

2 times a whole number, plus 1: that is the form of an odd number.

So n² + n + 1 is odd.

So n² + n + 1 is odd. ∎

The square at the end means: the proof is done.

5 Check

An example: n = 3.

It gives 13.

n = 3: 3² + 3 + 1 = 13

13 is 2 times 6, plus 1: odd, as the proof says.

Right!

13 = 2 × 6 + 1

Let's look back.

First: write it with letters.

Then: put them in.

Then: work it out.

Then: show the form.

Last: check.

Answer
2a² + 3a + 1 is a whole number

Your turn

Try this one yourself, then check:

Show the answer
n2 + n + 3 = 2(2a2 + 3a + 2) + 1

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