Welcome! Today we're exploring Divisibility Patterns in Consecutive Integers — one of those satisfying topics where a pattern you've always noticed finally gets a proper proof.

Here's a pattern you've probably noticed but never proved:

Pick any three consecutive numbers — say 7, 8, 9example — and exactly one of them is divisible by 3.

Always. No matter which three you pick.

We're going to prove this rigorously using the division lemma — no hand-waving, just solid reasoning.

Then we'll use it to crack a beautiful result:

Expression	Property
$n^3 - n$	Always divisible by 6key

For any positive integer $n$ . Always.

1. Checking divisibility for all three numbers in one case

Here's a pattern you've probably noticed but never proved:

Pick any three consecutive numbers — say 7, 8, 9 — and exactly one of them is divisible by 3.

Number line showing three consecutive integers 7, 8, 9 marked with dots, with 9 highlighted or circled to indicate divisibility by 3

Always. No matter which three you pick.

We're going to prove this rigorously.

The division lemma with divisor 3 gives us three cases for any integer $n$ :

n = 3q + r \quad \text{where } r = 0, 1, \text{ or } 2

Key discipline: In each case, you must check all three numbers — not just find the one that works.

📋 Given Info

Let's start with the first case: $r = 0$ , which means $n = 3q$ .

Our three consecutive integers are:

$n$
$n + 1$
$n + 2$

✍️ Question

Your turn 🧮

When $n = 3q$ , state the values of $n$ , $n + 1$ , and $n + 2$ .

For each one, say whether it is divisible by 3 or not — and if not, state what remainder it leaves.

How many of the three are divisible by 3?

When $n = 3q$ , let's check each of the three consecutive integers:

$n = 3 q$ — Is this divisible by 3? Yes — it's 3 times $q$ When something is 3 times anything, it's automatically divisible by 3. Remainder: 0.

$n + 1 = 3 q + 1$ — Is this divisible by 3? No — when you divide $3 q + 1$ by 3, you get quotient $q$ and remainder 1leftoverThat extra 1 has nowhere to go. The $+1$ prevents exact divisionAdding 1 to a multiple of 3 always leaves remainder 1.

$n + 2 = 3q + 2$ — Divisible by 3? No — remainder is 2leftover.

✍️ MCQ

Choose one

In Case 1 where

n = 3q

, how many of the three consecutive integers

n

,

n+1

,

n+2

are divisible by 3?

✓

So exactly oneAmong any three consecutive integers, exactly one is divisible by 3 ( $n$ ) is divisible by 3Divisible by 3, and the other two are notThe other two will have remainders 1 and 2.

Why do we need to check all three?

Because the claim isn't just "at least one is divisible by 3True even if multiple numbers work" — it's "exactly oneOne and only one — a stronger claim."

This is a crucial distinction! "At least one" would be satisfied even if two or all three were divisible. But we're proving something stronger.

If we only showed $n$ is divisible, someone could argue that maybe $n + 1$ or $n + 2$ is also divisible.

By checking each one's remainder, we rule that outRule out the other possibilities.

$n = 3 q$ → remainder 0divisibleThis one is divisible by 3 → divisible ✓
$n + 1 = 3 q + 1$ → remainder 1Remainder 1 means not divisible → not divisible ✗
$n + 2 = 3 q + 2$ → remainder 2Remainder 2 means not divisible → not divisible ✗

✍️ MCQ

Choose one

In Case 1 where

n = 3q

, we showed that

n

is divisible by 3, while

n + 1

and

n + 2

are not. What made us certain that

n + 1

and

n + 2

are NOT divisible by 3?

✓

2. The r = 1 and r = 2 cases — where the divisible number shifts

In the r = 0 case, it was $n$ itself that was divisible by 3.

In the other two cases, the divisible number shifts:

Sometimes it's $n + 1$
Sometimes it's $n + 2$

You need to track which one it is in each case.

You've already shown that when r = 0 ( $n = 3 q$ ), exactly $n$ is divisible by 3.

Now complete the proof for the other two cases.

Tip: Expressions like $3 q + 3$ can be simplified to show divisibility — factor out the 3!

✍️ Question

For r = 1 ( $n = 3 q + 1$ ) and r = 2 ( $n = 3 q + 2$ ), check all three numbers $n$ , $n + 1$ , $n + 2$ in each case.

Which number is divisible by 3 in each case? Show why by simplifying the expression.

Let's work through both cases carefully.

Case r = 1 (n = 3q + 1):

$n = 3 q + 1$ . Remainder when divided by 3: 1r=1Remainder 1 means not divisible. Not divisibleWe need remainder 0 for divisibility.
$n + 1 = 3 q + 2$ . Remainder: 2r=2Remainder 2 also means not divisible. Not divisibleNeed remainder 0, not 1 or 2.

$n + 2 = 3 q + 1 + 2 = 3 q + 3$ . Now, $3q + 3 = 3(q + 1)$ The key move is factoring out the 3. This IS3 times anything is divisible by 3 divisible by 3 — it's 3 times $(q + 1)$ Writing as 3 times something proves divisibility.

So n + 2winner!Exactly one of three is divisible — that's the pattern is the one divisible by 3.

✍️ MCQ

Choose one

In Case r = 1, we simplified

n + 2 = 3q + 3

to

3(q + 1)

. What algebraic property did we use?

✓

Case r = 2 (n = 3q + 2):

$n = 3 q + 2$ . Remainder: 2≠0. Not divisible.

$n + 1 = 3 q + 3 =$ $3(q + 1)$ Factor out the 3 and you're done. Divisible by 3.3 times anything means divisible by 3

✍️ Yes/No

Yes or No?

In Case r = 2, we found that

n + 1 = 3(q + 1)

is divisible by 3. Do we still need to check

n + 2

?

✓

$n + 2 = 3 q + 4$ . Now, 4 divided by 3 gives quotient 1 remainder 1. So $3 q + 4 = 3 q + 3 + 1 = 3 (q + 1) + 1$ . Remainder: 1. Not divisible.

So n + 1Not zero, not two — exactly one is the one divisible by 3.

✍️ Subjective

Your turn

Match each case to the number divisible by 3:

Case r = 0: Which is divisible?
Case r = 1: Which is divisible?
Case r = 2: Which is divisible?

Notice the pattern: as $r$ When remainder is zero, n itself works increases from 0 to 2, the divisible number shiftsThe position shifts depending on the remainder from $n$ N itself is divisible when remainder is zero to $n + 2$ Go to n plus 2 when remainder is 1 to $n + 1$ N plus 1 works when remainder is 2.

Number line showing three consecutive integers n, n+1, n+2 with arrows indicating which one is divisible by 3 for each case r=0,1,2

✍️ MCQ

Choose one

If

n

leaves remainder 1 when divided by 3, which of the three consecutive integers

n

,

n+1

,

n+2

is divisible by 3?

✓

In every case, exactly oneNot at least one, not at most one — exactly one of the three consecutive integers is divisible by 3.

The 'exactly one'keyThis precise wording matters for exams is confirmed by checking all threeWe verified this by examining every case — the non-divisible ones always have remainder 1 or 2Other two numbers can never be divisible by 3.

3. Factoring before case analysis

You've proved the consecutive integer result — that among any three consecutive integers, exactly one is divisible by 3. 🎯

Now we're going to use it. But first, there's a strategic move that makes the proof much cleaner.

Instead of cubing and doing messy algebra, you factor the expression first.

Consider the expression $n^3 - n$ .

Before doing any division lemma case analysis, there's an algebraic step that transforms this into something much easier to work with.

✍️ Question

Factor $n^3 - n$ completely.

What do you notice about the factors?

Why is this factored form useful for proving divisibility?

Let's work through the factoring step by step. Start by factoring out $n$ :

n^3 - n = n(n^2 - 1)

Now, $n^{2} - 1$ is a difference of squaresThis pattern appears frequently in factoring:

$n^{2} - 1 = (n - 1) (n + 1)$

✍️ MCQ

Choose one

What algebraic identity did we just use to factor

n^2 - 1

?

✓

So we get:

$n^{3} - n = n (n - 1) (n + 1) = (n - 1) \times n \times (n + 1)$

Notice something beautiful here — these are three consecutive integersThree numbers in a row: $(n-1)$ , $n$ , and $(n+1)$ consecutiveWhy the expression is always divisible by 6!

✍️ Yes/No

Yes or No?

Since

(n-1) \times n \times (n+1)

represents three consecutive integers, is this product always divisible by

3

?

✓

Number line showing three consecutive integers labeled n-1, n, n+1 with equal spacing marks between them

Look at these three factors: $n - 1$ , $n$ , $n + 1$ Three numbers sitting right next to each other. They're three consecutive integersThree numbers sitting right next to each other on the number line.

If $n = 5$ , you get $4, 5, 6$ .

If $n = 100$ , you get $99, 100, 101$ .

✍️ MCQ

Choose one

Since

n^3 - n = (n-1) \cdot n \cdot (n+1)

is the product of three consecutive integers, what can we conclude?

✓

Why does this help?

Because you've already proved that among any three consecutive integers, exactly one is divisible by 3This is the foundation that makes the whole proof work.

So the product $(n - 1)(n)(n + 1)$ The entire product automatically has 3 as a factor must contain a factor of 3.

This means $n^3 - n$ proven!The result we wanted to prove is always divisible by 3, no matter what integer $n$ you choose!

This is the 'factor before case analysis'Your go-to strategy for expressions like these strategy: instead of cubing $n = 3 q + r$ for each $r$ and wading through ugly algebra, you factor firstFactor into consecutive integers first and reduce the problem to a result you've already provedApply what you've already shown.

Much cleaner.

The key insight: $n^3 - n = n(n-1)(n+1)$ Recognize this pattern instantly — a product of three consecutive integerskey!Always include a multiple of 3. And we've already shown that any product of three consecutive integers is divisible by 3That's your shortcut!

✍️ MCQ

Choose one

Why is factoring

n^3 - n

into

(n-1)(n)(n+1)

more useful than directly substituting

n = 3q + r

?

✓

4. Proving divisibility by 6 — combining two arguments

You've factored $n^{3} - n$ into three consecutive integers and connected it to divisibility by 3.

But the claim is divisibility by 6, not just 3.

You need a second argument for divisibility by 2, and then a way to combine both.

What you've established so far:

You've shown that:

n^3 - n = (n-1)(n)(n+1)

This is the product of three consecutive integers.

You know this product is divisible by 3 because among any three consecutive integers, one is a multiple of 3.

✍️ Question

Now prove that $(n - 1) (n) (n + 1)$ is also divisible by 2.

Then explain why being divisible by both 2 and 3 means it's divisible by 6.

Divisibility by 2: Look at any two consecutive integersNumbers that come one after another with no gap, like $n - 1$ and $n$ . One is evenThey always switch back and forth and one is oddThe pattern never breaks — they alternatepatternThis alternating pattern is the key.

Number line showing consecutive integers n-1, n, n+1 with alternating even/odd labels, highlighting the pattern

If $n$ is even, then $n - 1$ is odd.

If $n$ is odd, then $n - 1$ is even.

So among $(n - 1)$ and $n$ , one is always evenGuaranteed, no exceptions.

✍️ MCQ

Choose one

If

n = 100

, which of the three consecutive integers

(n-1)

,

n

,

(n+1)

are even?

✓

So the product $(n - 1) (n) (n + 1)$ always contains at least one even factorEven one even factor is enough, which means the whole product is divisible by 2That's the rule you need to remember.

Divisibility by 3: You already proved that among three consecutive integersAny three in a row will work, exactly one is divisible by 3One of them is always divisible by 3, guaranteed. So the product $(n-1)(n)(n+1)$ productThis product automatically has 3 as a factor contains a factor of 33 is guaranteed to be a factor.

Combining them: The product is divisible by 2 AND by 3. Can we conclude it's divisible by 6?

✍️ Yes/No

Yes or No?

If a number is divisible by both 4 and 6, is it necessarily divisible by

4 \times 6 = 24

?

✓

Yes — but only because 2 and 3 are coprimeGCD is 1, no common factors (their GCD is 1They share no common factors, they share no common factors).

When two coprime numbers both divide a value, their product also divides itMultiply them and the product divides it too.

So $2 \times 3 = 6$ coprimeCheck if coprime, then conclude divisibility by 6 divides the product $(n - 1) (n) (n + 1)$ .

✍️ MCQ

Choose one

Why can't we conclude that divisibility by 4 and 6 implies divisibility by 24?

✓

Important: This reasoning wouldn't work for, say, 4 and 6.

A number can be divisible by both 4 and 6 without being divisible by 24.

Example: 12our counterexample(The number that breaks the pattern) is divisible by 4 ✓Passes the divisibility by 4 test and divisible by 6 ✓Passes the divisibility by 6 test

But 12 is NOT divisible by 24Fails divisibility by 24 ✗fails!Common factors cause the failure

✍️ MCQ

Choose one

Why doesn't divisibility by both

4

and

6

guarantee divisibility by

24

?

✓

The 'coprime' conditionThe essential condition for the rule is essential.

Since 2 and 3 share no common factorsGCD is 1, nothing in common (they're coprime), divisibility by both guarantees divisibility by their product 62×3Both dividing means product divides.

5. A different application — n squared minus 1 divisible by 8 for odd n

You've seen how factoring and consecutive integer reasoning prove divisibility by 6.

Here's a different application of the same toolkit — this time targeting divisibility by 8, which requires a clever choice of divisor.

📋 Given Info

Consider the claim:

If $n$ is an odd positive integer, then $n^2 - 1$ is divisible by 8.

You know from earlier work that every positive odd integer is of the form:

$n = 4q + 1$ , or
$n = 4q + 3$

✍️ Question

Using these two cases ( $n = 4q + 1$ and $n = 4q + 3$ ), compute $n^2 - 1$ for each and show that both results are divisible by 8.

Let's expand both cases step by step.

Case $n = 4 q + 1$ :

$n^{2} = (4 q + 1)^{2} =$ $(4q)^2 + 2(4q)(1) + 1^2$ You'll use this identity again and again $= 16 q^{2} + 8 q + 1$

$n^{2} - 1 = 16 q^{2} + 8 q + 1 - 1 =$ $16q^2 + 8q$

✍️ MCQ

Choose one

Looking at

16q^2 + 8q

, what is the greatest common factor you can pull out from both terms?

✓

Factor out 8When you factor out 8 and see 8 times something: $16 q^{2} + 8 q = 8 q (2 q + 1)$

This is clearly divisible by 8 — it's 8 times somethingkey!You've proven divisibility by 8.

Case $n = 4q + 3$ :

$n^{2} = (4 q + 3)^{2} = (4 q)^{2} + 2 (4 q) (3) + 3^{2} =$ $16q^2 + 24q + 9$

$n^2 - 1$ $= 16 q^{2} + 24 q + 9 - 1 =$ $16q^2 + 24q + 8$

✍️ Yes/No

Yes or No?

Look at

16q^2 + 24q + 8

. Is 8 a common factor of all three terms?

✓

Factoring out 88 is the common factor in every term:

$16 = 8 \times 2$ , $24 = 8 \times 3$ , $8 = 8 \times 1$

So: $n^{2} - 1 =$ $8(2q^2 + 3q + 1)$ 8 is factorOnce you see 8 as a common factor, divisibility is proven

✓ Divisible by 8That's the whole point of this factoring step

Why divisor 4 instead of divisor 2?

Notice why we used divisor 4 (giving forms $4 q + 1$ and $4 q + 3$ for odd integers) rather than divisor 2 (giving $2 q + 1$ ).

This is a crucial strategic choiceThe divisor determines what factors appear — the divisor you pick determines what factors naturally appear in your algebra.

What happens with divisor 2:

With divisor 2, you'd get: $n^{2} - 1 = (2 q + 1)^{2} - 1 = 4 q^{2} + 4 q = 4 q (q + 1)$

This only shows divisibility by 4not 8!Only divisible by 4, not the 8 we need, not by 8.

See the problem? The factor of 8 just doesn't appear naturallyYou'd need extra work to find divisibility by 8.

✍️ MCQ

Choose one

When we used divisor 2, we got

n^2 - 1 = 4q(q+1)

. This directly shows divisibility by which number?

✓

What happens with divisor 4:

The larger divisor gives terms involving $16q^2$ These coefficients are all multiples of 8 and $8q$ These coefficients are all multiples of 8, which are both divisible by 8.

Key insight:Divisor 4 makes the factor of 8 appear naturally Choosing the right divisor is what makes the target factor appear naturally.

When you want to prove divisibility by 8, using divisor 4 gives you coefficients like 16 and 8÷816, 8, 24 are all divisible by 8 — both multiples of 8. The factor of 8 just falls out of the algebraNo extra effort needed to find divisibility by 8 without any extra cleverness needed.