Equivalence Classes of the Relation a² − b² Divisible by 5 – JEE Main 2026 PYQ

Nishant Kumar Gupta

Former Faculty · Allen · Aakash · Narayana

Maths mentor behind PadhoLikhoJEE and Quantum Chemistry Classes (Arrah, Bihar). Teaches the “why” behind every JEE trap — not just the trick.

Quick Summary

Question Type:

Statement I / Statement II (concept + counting)

Chapter:

Relations and Functions

Difficulty:

⭐⭐ Easy

Time to Solve:

60–90 seconds

Key Concept:

aRb really means a² ≡ b² (mod 5); squares mod 5 take only 3 values {0, 1, 4}, so there are 3 classes — and R is reflexive, symmetric, transitive

Correct Answer:

Option (3) – Both I and II are true

Why:

R is an equivalence relation, and since a² mod 5 ∈ {0, 1, 4}, the elements split into exactly 3 classes — not 5

The Question

JEE Main 2026 (8 April – Evening Shift) – Relations and Functions

Consider the relation R on the set of integers ℤ defined by
(a, b) ∈ R if and only if a² − b² is divisible by 5.

I. The number of equivalence classes of R is 3.
II. R is an equivalence relation.

1. (1) Only I is true
2. (2) Only II is true
3. (3) Both I and II are true ✓
4. (4) Neither I nor II is true

Quick Answer

Correct Option: (3) – Both I and II are true

Reasoning: one congruence rewrite settles both statements:

• Read the condition as congruence: 5 | (a² − b²) means exactly a² ≡ b² (mod 5) — R compares squares, not the numbers themselves
• Equivalence check (II): a² − a² = 0 (reflexive); a² − b² and b² − a² differ by sign (symmetric); the sum of two multiples of 5 is a multiple of 5 (transitive) → II is true
• Counting classes (I): testing residues 0, ±1, ±2 gives a² ≡ 0, 1, 4 (mod 5) — only 3 possible square values, hence 3 classes → I is true
• The trap: ℤ has 5 residue classes, but ±1 collapse into one class (both square to 1) and ±2 collapse into another (both square to 4) ✓

Understanding the Concept

From Divisibility to Congruence

The definition “(a, b) ∈ R iff 5 divides a² − b²” is precisely the statement that a² and b² leave
the same remainder on division by 5. Whenever a relation is built from divisibility of a
difference, the fastest route is to translate it into modular notation, because congruence
automatically bundles the three defining properties of an equivalence relation: same remainder
as itself (reflexivity), remainder comparison works in both directions (symmetry), and two
numbers sharing a remainder share it with each other (transitivity). This single observation
proves Statement II in one line instead of three separate paragraphs.

Why the Classes Are Classified by Squares

Two integers a and b are related exactly when their squares agree mod 5. So the class of an
element is decided by the value of a² mod 5, not by a mod 5. Computing a² mod 5 for the five
residue classes of ℤ mod 5 gives the complete picture:

The residues 1 and 4 (i.e. ±1) both square to 1, and the residues 2 and 3 (i.e. ±2) both square
to 4. That collapsing of ± pairs is exactly why five residue classes shrink to three
square-classes. It also explains the structure of each class: the class of 0 is the set of
multiples of 5, the “1-class” contains all integers ≡ ±1 (mod 5), and the “4-class” contains
all integers ≡ ±2 (mod 5).

The General Picture (Worth Memorising)

For an odd prime p, the nonzero squares mod p come in ± pairs, so there are (p − 1)/2 nonzero
square values and (p + 1)/2 classes in total (including the 0-class). For p = 5 this predicts
(5 + 1)/2 = 3 classes — matching Statement I instantly. The same machinery shows why the
famous relation “a² − b² divisible by 8” splits ℤ into classes governed by n² mod 8 ∈
{0, 1, 4}, and why “n² − m² divisible by 4” merges odds and evens in a particular way. JEE
recycles this template across many moduli, so the method matters far more than the number 5.

Detailed Step-by-Step Solution

Final Answer

Essential Facts for This Topic

The Working Toolkit

1. Divisibility ⟺ congruence: n | (x − y) ⟺ x ≡ y (mod n) — the standard first move for every “difference divisible by” relation
2. Squares mod 5: {0, 1, 4} — memorise the table 0², 1², 2², 3², 4² ≡ 0, 1, 4, 4, 1
3. Class structure: [0] = 5ℤ; [1] = {…, −4, −1, 1, 4, 6, 9, …} (a ≡ ±1); [4] = {…, −3, −2, 2, 3, 7, 8, …} (a ≡ ±2)
4. Counting rule for odd primes: number of classes of “a² ≡ b² (mod p)” = (p + 1)/2 — for p = 5 that is 3, for p = 7 it is 4 (squares 0, 1, 2, 4)
5. Classic sibling: n² mod 4 ∈ {0, 1} separates ℤ into odds and evens; n² mod 8 ∈ {0, 1, 4} gives three classes for the “divisible by 8” version

Common Mistakes to Avoid

Key Concept Summary

What You Must Remember

1. Translate first: “difference divisible by n” ≡ “same remainder mod n” — then the equivalence properties are automatic
2. Squares mod 5 = {0, 1, 4}: three square values ⟹ three classes, no matter how many residue classes ℤ has
3. ± collapse: a and −a always share a class in square-comparison relations — that is what shrinks 5 to 3
4. Equivalence proof = 3 one-liners: zero difference, sign flip, sum of multiples — write all three
5. General count: for odd prime p, classes = (p + 1)/2; check p = 5 → 3, p = 7 → 4

The Golden Rule for Square-Comparison Relations

“The relation compares squares, not numbers — classify by a² mod n, and remember a and −a always live in the same class.”

Frequently Asked Questions

Q1: How many equivalence classes does R have and what are they?

A: Exactly three. The class of multiples of 5 (where a² ≡ 0), the class of integers congruent to ±1 mod 5 (where a² ≡ 1), and the class of integers congruent to ±2 mod 5 (where a² ≡ 4). Every integer lands in exactly one of these three classes.

Q2: Why is the answer 3 classes and not 5?

A: Because the relation compares a² with b², and squares mod 5 take only the values 0, 1 and 4. The five residue classes of ℤ collapse in pairs: +1 and −1 both square to 1, and +2 and −2 both square to 4. Only the zero class stands alone, giving 1 + 1 + 1 = 3 classes.

Q3: How do you quickly prove R is an equivalence relation?

A: Reflexive: a² − a² = 0 is divisible by 5. Symmetric: if 5 divides a² − b² it also divides the negation b² − a². Transitive: a² − c² = (a² − b²) + (b² − c²) is a sum of two multiples of 5, hence a multiple of 5. Three one-line checks complete the proof.

Q4: Is there a shortcut to count classes for similar questions with other moduli?

A: Yes — compute the set of square values {a² mod n}. For an odd prime p the count of classes is (p + 1)/2 because nonzero squares pair up as ±a. For p = 7, squares are 0, 1, 2, 4 giving 4 classes; for mod 8 the squares are 0, 1, 4 giving 3 classes.

Q5: What is the equivalence class of 0?

A: The class of 0 is the set of all multiples of 5: {…, −10, −5, 0, 5, 10, …}. These are exactly the integers whose square is ≡ 0 (mod 5), i.e. the integers themselves divisible by 5.

Prerequisites to Solve This Question

Before attempting this problem, you should be comfortable with:

1. Modular arithmetic: computing a² mod 5 for the five residue classes and reading “divisible” as congruence
2. Equivalence relation definition: reflexive, symmetric, transitive — with the standard one-line proofs
3. Equivalence classes: a class is the set of everything related to a given element, and classes partition ℤ
4. Division algorithm: every integer is uniquely 5q + r with r ∈ {0, 1, 2, 3, 4}
5. Basic set notation: describing classes like {a ∈ ℤ : a ≡ ±1 (mod 5)} cleanly

After Solving This, You Can:

• Count equivalence classes for “a² − b² divisible by 7” in under a minute (answer: 4, squares 0, 1, 2, 4)
• Handle the mod 8 version and explain why n² mod 8 never equals 2 or 6
• Prove or disprove equivalence for look-alike relations such as “a·b divisible by 5” (fails reflexivity at a = 1)
• Convert any “difference divisible by n” relation into congruence language automatically
• Predict the class count for prime moduli using the (p + 1)/2 rule

Study Tips for This Topic

For JEE Main:

1. Memorise the square tables: squares mod 5 = {0, 1, 4} and mod 8 = {0, 1, 4} — these two cover most JEE variants of this question
2. Write the three one-liners: reflexivity, symmetry, transitivity proofs take 15 seconds each; skipping them risks losing the “R is an equivalence relation” statement marks in subjective re-tests
3. Budget 60–90 seconds: this is a two-statement question — read the relation as congruence, count square values, tick Both true

Common JEE Variants:

• “a − b is divisible by 5” — the plain congruence version with 5 classes (no ± collapse)
• “a² − b² is divisible by 8” — classes counted by n² mod 8 ∈ {0, 1, 4}
• “a + b is divisible by 3” — classes pair up as negatives; good symmetry practice
• “a·b is even” — test which of the three properties fails (reflexivity) instead of assuming equivalence
• Find the class [a] explicitly and verify two given integers lie in the same class

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐ (2/5) – Easy (two mechanical steps: rewrite as congruence, count square values; the only bite is resisting the “5 classes” instinct)

JEE Main Frequency: High — equivalence relations with divisibility conditions appear in nearly every JEE Main session in the Relations and Functions slot

JEE Advanced Frequency: Low to Medium — Advanced prefers functional relations or matrix-based equivalence definitions, but the modular counting idea reappears

Topic Importance: Very High — equivalence classes are the backbone of Relations and Functions and feed directly into functions and number-theory topics

Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 2 Sep 2026 ·
Question Source: JEE Main 2026 (8 April – Evening Shift) Previous Year Question ·
Topic: Relations and Functions – Equivalence Classes via Squares Modulo 5