In short

A divisibility statement about numbers is almost always a factorisation statement about expressions in disguise. The key bridge: a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}), so a^n - b^n is always divisible by a - b — for any integers a and b. Want to show 7^n - 1 is divisible by 6? Write it as 7^n - 1^n and read off the factor 7 - 1 = 6. Most CBSE, JEE, and Olympiad "prove divisible by k" problems collapse to spotting this one move.

You open a problem and it says: Show that 7^{100} - 1 is divisible by 6. You stare at 7^{100}, a number with 85 digits, and panic — there is no way you are computing that. Then you remember an identity from this very chapter: a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1}). Set a = 7, b = 1, n = 100. The first factor is 7 - 1 = 6. Done. The 85-digit number is 6 times something — you do not need to know what something is, you just need to know it exists.

That move — turning a number-divisibility question into a factorisation question — is the whole game. This article is about how the identities you already know become weapons in number theory, why "divisible by k" and "has k as a factor" mean exactly the same thing, and how a CBSE-style induction problem about divisibility by 64 secretly relies on one of these identities for its inductive step.

The core principle: a factor in algebra is a factor in arithmetic

Every algebraic identity that writes one expression as a product of two others is, secretly, a divisibility theorem.

The headliner is the difference of n-th powers:

a^n - b^n \;=\; (a - b)\bigl(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}\bigr)

This holds for every positive integer n and every pair of values a, b. You can verify n = 2 (a^2 - b^2 = (a-b)(a+b)), n = 3 (a^3 - b^3 = (a-b)(a^2 + ab + b^2)), and so on, by direct multiplication. The pattern continues forever.

Why does an algebraic factor become a numeric factor? Because when you plug specific integers for a and b, both sides of the identity become specific integers — and the right side is literally (a-b) times an integer (the long second bracket evaluates to a whole number when a and b are whole). So the left side is an integer multiple of (a-b). That is exactly what "divisible by a-b" means.

The bridge from algebra to arithmetic is one sentence: if an expression factors as P \cdot Q over the integers, then plugging in any integer values gives a number that is divisible by both P and Q. Algebra hands you the factorisation; arithmetic reads off the divisibility for free.

Factorisation in algebra implies divisibility in arithmeticA two-box flow diagram. Left box, labelled algebra, contains the identity a to the n minus b to the n equals open paren a minus b close paren times a long second factor. An arrow labelled plug in integers leads to the right box, labelled arithmetic, which contains the statement that the resulting number is divisible by a minus b. A caption underneath says: a factor in the expression becomes a factor in the number. algebra (identity) aⁿ − bⁿ = (a − b) · (long factor) true for all values plug in integers arithmetic (divisibility) aⁿ − bⁿ is divisible by (a − b) for any integers a, b, n A factor in the expression becomes a factor in the number.
The same identity, read two ways. As algebra, it is a recipe for rewriting one expression as a product. As arithmetic, plugging in any integers turns that product into a divisibility statement — for free.

Common patterns you will meet again and again

A handful of identities, applied in this divisibility key, cover most of the textbook problems and a healthy fraction of competition ones.

Pattern 1 — a^n - 1 is divisible by a - 1. Special case of a^n - b^n with b = 1. So 7^n - 1 is divisible by 7 - 1 = 6 for every positive integer n. So is 11^n - 1 divisible by 10, and 3^n - 1 divisible by 2. The number a chooses the divisor.

Pattern 2 — a^n - b^n is divisible by a - b. The general form. So 25^n - 24^n is divisible by 1 (trivially); 25^n - 16^n is divisible by 9; 5^7 - 3^7 is divisible by 2. Whenever you see two powers being subtracted, the difference of bases is a guaranteed factor.

Pattern 3 — a^3 + b^3 + c^3 - 3abc is divisible by a + b + c. This is the cubic identity a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) in disguise. Plug in any three integers and the resulting number is automatically divisible by their sum. Try a = 2, b = 3, c = 5: the expression is 8 + 27 + 125 - 90 = 70, and indeed 70 is divisible by 2 + 3 + 5 = 10.

Pattern 4 — n^2 - n is divisible by 2. Factor it: n^2 - n = n(n - 1). Two consecutive integers; one of them is even. So the product is even. Why: among any two consecutive integers exactly one is divisible by 2, so the product picks up that factor of 2 no matter what n is.

Pattern 5 — n^3 - n is divisible by 6. Factor it: n^3 - n = n(n^2 - 1) = n(n-1)(n+1) = (n-1) \cdot n \cdot (n+1). Three consecutive integers. Among any three consecutive integers, one is divisible by 3, and at least one is divisible by 2. So the product picks up both a 2 and a 3, giving 6.

This last one is a tiny gem — a question that looks like it needs cubes and high-powered algebra dissolves into "three consecutive numbers, count the small factors."

Lock-in: show $25^7 - 24^7$ is divisible by $1$

This is deliberately silly — the divisor is 25 - 24 = 1, and every integer is divisible by 1. But the point is to see the technique with no distraction.

By the difference-of-powers identity, 25^7 - 24^7 = (25 - 24) \cdot K for some positive integer K, so 25^7 - 24^7 = 1 \cdot K = K, which is trivially a multiple of 1. The same template — write the big number as (a - b) times something — will solve every problem in this article. Get used to its shape now and you save thinking later.

Apply twice: show $a^4 - b^4$ is divisible by $a^2 - b^2$

Two routes; both arrive at the same answer.

Route 1 (general identity). Take a^n - b^n with n = 4 and rewrite using a^2 and b^2 as the bases: a^4 - b^4 = (a^2)^2 - (b^2)^2 = (a^2 - b^2)(a^2 + b^2). Done — the factor a^2 - b^2 is right there.

Route 2 (difference of squares twice). a^4 - b^4 = (a^2 - b^2)(a^2 + b^2), then push further if you like: a^2 - b^2 = (a - b)(a + b), giving a^4 - b^4 = (a - b)(a + b)(a^2 + b^2). Why three factors? Because for n = 4, the expression has more structure than for prime n — every divisor of 4 (namely 1, 2, 4) gives a factorisation route. So a^4 - b^4 is divisible by a - b, by a + b, and by a^2 - b^2. Algebra is generous when the exponent has many divisors.

So plugging in integers: a = 5, b = 3 gives 625 - 81 = 544, which is divisible by 5^2 - 3^2 = 16. Check: 544 / 16 = 34. Clean.

A worked CBSE-style problem: divisible by 64

Here is the kind of problem that shows up in NCERT exemplars and CBSE board papers.

Problem. Show that 9^{n+1} - 8n - 9 is divisible by 64 for every positive integer n.

The standard tool is mathematical induction, but the inductive step uses an algebraic-identity manipulation. That is the punchline of this article — induction is the scaffolding, identity is the brick.

Base case (n = 1). 9^{2} - 8 \cdot 1 - 9 = 81 - 8 - 9 = 64. And 64 is divisible by 64. The base case checks.

Inductive step. Assume the statement holds for some n = k: that is, 9^{k+1} - 8k - 9 = 64m for some integer m. Show it then holds for n = k + 1: that is, 9^{k+2} - 8(k+1) - 9 is divisible by 64.

Start from the target and rewrite using the inductive hypothesis:

9^{k+2} - 8(k+1) - 9 \;=\; 9 \cdot 9^{k+1} - 8k - 17

From the hypothesis, 9^{k+1} = 64m + 8k + 9. Substitute:

9 \cdot (64m + 8k + 9) - 8k - 17 \;=\; 576m + 72k + 81 - 8k - 17 \;=\; 576m + 64k + 64

Factor out 64:

576m + 64k + 64 \;=\; 64(9m + k + 1)

Which is divisible by 64, as required. Why this works: the substitution step replaced one 9^{k+1} with the form 64m + 8k + 9, and then ordinary algebra collected like terms. Every term turned out to be a multiple of 64 — the 576m from 9 \cdot 64m, the 64k from 72k - 8k, and the 64 from 81 - 17. The inductive hypothesis was the only thing that guaranteed the 64m piece existed; the rest was algebraic identity work — distribution, combining like terms, factoring out the common 64.

By induction, 9^{n+1} - 8n - 9 is divisible by 64 for every positive integer n. \blacksquare

The induction frame is the same in every problem of this shape; the algebra inside the inductive step is where you actually have to think. That algebra is just identity work.

JEE-style: show $5^n + 2 \cdot 3^{n-1} + 1$ is divisible by $8$ for every positive integer $n$

Same plan — induction, with identity-style algebra in the inductive step.

Base case (n = 1). 5 + 2 \cdot 3^0 + 1 = 5 + 2 + 1 = 8. Divisible by 8. Check.

Inductive step. Assume 5^k + 2 \cdot 3^{k-1} + 1 = 8m for some integer m. Show 5^{k+1} + 2 \cdot 3^k + 1 is divisible by 8.

Write the target by pulling factors out:

5^{k+1} + 2 \cdot 3^k + 1 \;=\; 5 \cdot 5^k + 6 \cdot 3^{k-1} + 1

The trick: split 5 \cdot 5^k as 5^k + 4 \cdot 5^k, and 6 \cdot 3^{k-1} as 2 \cdot 3^{k-1} + 4 \cdot 3^{k-1}:

= \underbrace{\bigl(5^k + 2 \cdot 3^{k-1} + 1\bigr)}_{= \,8m \text{ by hypothesis}} + 4 \cdot 5^k + 4 \cdot 3^{k-1} \;=\; 8m + 4\bigl(5^k + 3^{k-1}\bigr)

Now you need 5^k + 3^{k-1} to be even, so that 4 \cdot \text{(even)} = 8 \cdot \text{(integer)}. And it is: 5^k is odd (any power of an odd number is odd) and 3^{k-1} is odd, so their sum is even. Why split that way? You want to expose the inductive hypothesis inside the new expression. Adding and subtracting cleverly to make the previous case visible is the universal induction-with-identity trick. Once the hypothesis is visible, it gets replaced by 8m and the rest of the algebra has to give a multiple of 8 for the proof to close.

So 5^{k+1} + 2 \cdot 3^k + 1 = 8m + 8 \cdot \tfrac{5^k + 3^{k-1}}{2}, which is a multiple of 8. By induction, the result holds for every n \geq 1. \blacksquare

Why this matters past school

Indian competitive maths leans heavily on these moves. JEE Main and Advanced both regularly ask divisibility-by-induction problems, and the inductive step is always powered by an algebraic identity. The Indian National Mathematical Olympiad (INMO) and the Regional Mathematical Olympiad (RMO) lean even harder on the difference-of-powers identity — it is the standard opening move for problems of the shape "show a^n - b^n has a particular prime factor for all n."

A famous Olympiad classic: prove that n^5 - n is divisible by 30 for every integer n. Try it — factor n^5 - n = n(n^4 - 1) = n(n^2-1)(n^2+1) = (n-1)n(n+1)(n^2+1) and chase the factors 2, 3, and 5. Three consecutive integers handle 2 and 3; the 5 comes from a small case argument on residues. The whole proof is identity work plus elementary counting. No fancy machinery.

The lesson is simple. Whenever a problem asks you to prove a divisibility statement, your first instinct should not be to compute. It should be to factor. The identity book you spent a chapter learning is also a number-theory book, written in disguise.

The takeaway

Divisibility statements about numbers are factorisation statements about expressions. The identity a^n - b^n = (a - b)(a^{n-1} + \cdots + b^{n-1}) alone unlocks an enormous fraction of school and competition problems — anything of the shape "a^n - b^n is divisible by a - b" reads off in one line. Cubic and consecutive-integer identities cover most of the rest. CBSE-style problems wrap this in induction; the inductive step is still identity work, just with an extra hypothesis to feed in.

One question to ask before any divisibility problem: can I write this expression as a product where one factor is the divisor I want? If yes — and an identity usually says yes — you are done before you start computing.

References

  1. NCERT Class 11 Mathematics, Chapter 4: Principle of Mathematical Induction — the canonical CBSE treatment, including divisibility-by-induction worked examples.
  2. NCERT Class 9 Mathematics, Chapter 2: Polynomials — where the identities a^3 + b^3 + c^3 - 3abc = (a+b+c)(\ldots) and the cubic factorisations are introduced.
  3. Art of Problem Solving — Difference of powers — competition-style writeup of the a^n - b^n family and its applications.
  4. HBCSE — Indian Mathematical Olympiad past papers — INMO and RMO problem archives, where divisibility-by-identity is a recurring theme.
  5. Burton, Elementary Number Theory — standard reference for the algebra-arithmetic bridge underpinning these problems.