INMO Number Theory Questions 2026 with Solutions

INMO Number Theory Questions 2026 will help you understand the depth of important topics such as Divisibility, Prime Numbers, Perfect Squares, and the sum of digits function S(n). The Homi Bhabha Centre for Science Education (HBCSE) conducts the INMO to select talent for the International Mathematical Olympiad (IMO) training camp. Number Theory often requires elegant logical derivations rather than just computation. To help aspirants excel, here is a compiled list of INMO Number Theory Most Expected Questions based on the latest training camp modules and DPPs.

Also Read:

• INMO Syllabus 
• How to Prepare for INMO

INMO Number Theory Exam Overview 2026

The INMO 2026 exam consists of high-level subjective problems. Number Theory questions typically require rigorous proofs involving properties of integers. Candidates must focus on step-by-step logic and clear mathematical induction to score full marks in this competitive arena.

INMO Number Theory PDF Resources

To aid your preparation, we have categorized the Daily Practice Problems (DPP) from the INMO Training Camp. These documents contain core problems and detailed solutions essential for the 2026 cycle.

 INMO Previous Year Question Paper

INMO Number Theory Important Topics

Based on the INMO Training Camp 2026 curriculum, the following topics are crucial for a robust foundation:

• Divisibility and Primality: Properties of prime numbers and divisibility rules.

• Perfect Squares and Cubes: Finding integer solutions where expressions result in perfect powers.

• Diophantine Equations: Solving equations like p^m − n^3 = 8 or 2^m = 7n^2 + 1 for integer solutions.

• Sum of Digits: Problems involving S(n) and modular arithmetic (mod 9).

• Divisor Functions: Analyzing tau(n) and subsets of divisors with specific remainders.

INMO Number Theory Questions 2026 with Solutions

Below are some of the INMO Number Theory important questions curated from the 2026 training camp practice sets.

Q1: Find all positive integers n for which 3^4 + 3^5 + 3^6 + 3^7 + 3^n is a perfect square.

Answer: Factoring the first terms: 3^4(1 + 3 + 9 + 27 + 3^(n−4)) = 3^4(40 + 3^(n−4)). For this to be a square, 40 + 3^(n−4) must be a square s^2. Case analysis yields n = 6 and n = 8.

Q2: Find all integers n for which n^2 + 2^n is a perfect square.

Answer: For n < 4, n = 3 gives 9 + 8 = 17 (no), but n = 0 gives 0 + 1 = 1 (yes) and n = 6 gives 36 + 64 = 100 (yes). For odd n ≥ 1, let n = 2k + 1; analysis of (m − n)(m + n) powers of 2 shows no solutions for odd n > 0. The solutions are n ∈ {0, 6}.

Q3: Determine all pairs (m, n) of positive integers such that 2^m = 7n^2 + 1. 25

Answer: Note 2^m ≡ 1 (mod 7), so m = 3k. Then 2^(3k) − 1 = (2^k − 1)(2^(2k) + 2^k + 1) = 7n^2. Factoring A and B where gcd(A, B) ∈ {1, 3} leads to the solution (m, n) = (3, 1) and (9, 9).

Q4: Determine all positive integers n such that there exist a, b satisfying S(a) = S(b) = S(a + b) = n.

Answer: Since a ≡ S(a) (mod 9), we have a ≡ n, b ≡ n, and a + b ≡ n (mod 9). This implies 2n ≡ n (mod 9), so n must be a multiple of 9.

Q5: Determine all pairs (p, q) of prime numbers for which p^(q−1) + q^(p−1) is a perfect square.
 

Answer: If p, q are both odd, p^(q−1) + q^(p−1) is even; x^2 + y^2 ≡ 2 (mod 4) is impossible for a square unless p = q. If p = q = 2, 2^1 + 2^1 = 4 = 2^2 (solution). If q = 2, p^1 + 2^(p−1) must be a square; trial shows (p, q) = (3, 2) is not a solution. The only solution is (2, 2).

Q6: Determine all pairs (a, b) of positive integers such that a^2b divides b^2 + 3a.

Answer: Let b^2 + 3a = k a^2b. This implies a divides b^2 and b divides 3a. Testing mn = 9 where b^2m = 9a and an = b^2 leads to solutions (1, 1), (1, 3), (3, 3).

Q7: Let a, b be positive integers such that a > b and a − b = 5b^2 − 4a^2. Prove a − b is a square.
 

Answer: Rewrite as (a − b)(1 + 4(a + b)) = b^2. Proving gcd(a − b, 1 + 4(a + b)) = 1 ensures both terms are perfect squares.

Q8: Prove that there are no positive integers m, n such that 5m^3 = 27n^4 − 2n^2 + n.

Answer: Analyzing the equation modulo 5 or using bounds on the polynomial growth shows that the left and right sides cannot coincide for positive integers.

Q9: Find all integral values of the fraction τ(10n) / τ₁(10n) where τ₁(n) counts divisors ≡ 1 (mod 3).

Answer: The fraction can take the value 2(a + 1), meaning any even positive integer is a possible value.

Q10: Prove that for every n there exist a, b such that n divides 4a^2 + 9b^2 − 1.

Answer: If n = 2k + 1, let a = k, b = 0; then 4k^2 − 1 = (2k − 1)(2k + 1), which is divisible by n. Similar constructions exist for even n.

Q11: Determine all (a, b) such that overline(a.b) · overline(b.a) = 13.

Answer: The condition is equivalent to (a + b / 10^k)(b + a / 10^m) = 13. For digits, ab · ba ≈ 1300 leads to (a, b) = (2, 5) and (5, 2).

Q12: Prove that if a, b have different parity, there exists c such that ab + c, a + c, b + c are squares.

Answer: Define c = (1 + a^2 + b^2 − 2ab − 2a − 2b) / 4 or a similar quadratic form; specific choices of c satisfy the requirement.

Q13: Prove that the sequence a_k = floor(2^k / k) contains infinitely many odd numbers.

Answer: Let k = 3 · 4^i. The fraction 2^k / k leaves a remainder that, when floored, consistently yields an odd integer.

Q14: If n | p − 1 and p | n^3 − 1, prove 4p − 3 is a square.

Answer: Since p | (n − 1)(n^2 + n + 1) and p > n, p must divide n^2 + n + 1. If p = n^2 + n + 1, then 4p − 3 = 4n^2 + 4n + 1 = (2n + 1)^2.

Q15: Determine all positive integers x, y such that x(x^2 + 19) = y(y^2 − 10).

Answer: Rewriting as d^2(b^3 − a^3) = 19a + 10b and testing divisors of 29 leads to solutions (x, y) = (2, 3).

Q16: Determine triplets (p, m, n) such that p is prime and p^m − n^3 = 8.

Answer: p^m = (n + 2)(n^2 − 2n + 4). Both factors must be powers of p. Solutions are (p, m, n) ∈ {(3, 2, 1), (2, 4, 2)}.

Q17: Determine triplets (p, m, n) such that p is prime and 2^m p^2 + 1 = n^5.

Answer: Factoring n^5 − 1 = (n − 1)(n^4 + n^3 + n^2 + n + 1). Since p is odd, analysis of the power of 2 shows n − 1 = 2^m.

Q18: Determine all n for which there exists d | n such that dn + 1 | d^2 + n^2.

Answer: If k = (d^2 + n^2) / (dn + 1), analysis shows k must be a perfect square. For d = 1, n + 1 | n^2 + 1 gives n = 1.

Q19: Prove that 51^k − 17 is divisible by 2^n for some k.

Answer: Using induction on n, for n = 1, 51 − 17 = 34 is even. The step n → n + 1 follows from 51^(2^(n−2)) ≡ 1 (mod 2^n).

Q20: Prove there exist a₁, …, aₙ such that (a_i + a_j) / (a_j − a_i) is always an integer.

Answer: Construct the set using induction: if {a₁, …, aₙ} is good, then {L, L + a₁, …, L + aₙ} is good for a sufficiently large L.

Preparation Tips for INMO Number Theory 2026

To succeed in the INMO Number Theory section, candidates should follow a rigorous approach:

Master the Basics: Ensure deep understanding of modular arithmetic, Fermat’s Little Theorem, and Euler’s Totient Theorem before complex lemmas.

Drafting Proofs: In INMO, the method is as important as the result. Practice writing full formal proofs daily.

Factorization Techniques: Many problems reduce to factoring expressions like x^n ± y^n and analyzing the factors.

Modular Constraints: Always check small moduli (3, 4, 7, 8, 9) to eliminate cases or find patterns in perfect powers.

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