Number Theory in 1 Shot ISI–CMI Maths Mastery 2026

ISI–CMI mathematics demands sharp problem-solving skills and deep conceptual clarity. Number Theory stands out as one of the most scoring yet challenging areas, where small insights can unlock complex problems.

Focus on building intuition through carefully chosen problems. By exploring trajectories, modular arithmetic, and equation-based reasoning, you’ll learn how to approach unfamiliar questions with confidence and precision.

Understanding The Game And Trajectories

The core of the problem introduces a game played with a positive integer a > 1.

Game Rules:

1. Starting Point: Begin with any positive integer a > 1.

2. Rule 1 (Perfect Square): If a is a perfect square, replace a with its square root (√a).

3. Rule 2 (Non-Perfect Square): If a is not a perfect square, replace a with a + 3.

4. Repetition: Repeat the procedure with the newly generated positive integer.

The sequence of numbers generated by applying these rules forms the Trajectory of A.

Examples of Trajectories:

• Example 1: Starting with a = 16
  16 → √16 = 4
  4 → √4 = 2
  2 → 2 + 3 = 5
  5 → 5 + 3 = 8
  … (This process continues indefinitely)
  Trajectory: {16, 4, 2, 5, 8, ...}

• Example 2: Starting with a = 7
  7 → 7 + 3 = 10
  10 → 10 + 3 = 13
  13 → 13 + 3 = 16
  16 → √16 = 4
  4 → √4 = 2
  2 → 2 + 3 = 5
  … (This sequence eventually merges with Example 1)
  Trajectory: {7, 10, 13, 16, 4, 2, 5, 8, ...}

• Example 3: The *3-6-9 Trajectory (A Special Case)*
  Starting with a = 3:
  3 → 3 + 3 = 6
  6 → 6 + 3 = 9
  9 → √9 = 3
  This sequence forms a loop: {3, 6, 9, 3, 6, 9, ...}. The set {3, 6, 9} represents this finite trajectory.

Main Problem Statement (CMI 2024 Question)

The central question is: Which numbers have a finite trajectory?

This involves finding the set S where n is the smallest number in a finite trajectory.

Sub-Problem A: Show That There Is No Trajectory Of Cardinality One Or Two.

Cardinality refers to the number of distinct elements in a trajectory set.

Crucial Initial Claim: If n is the smallest number in a trajectory S, then n cannot be a perfect square. If n were a perfect square, √n would be in the trajectory, and since n > 1, √n < n, contradicting n being the smallest.

Case 1: Cardinality = 1

Assume S = {n}. This means n must map to n.

• If n is a perfect square, √n = n implies n = 1, which is not allowed (a > 1).

• If n is not a perfect square, n + 3 = n implies 3 = 0, which is impossible.
  Conclusion: A trajectory of cardinality 1 is not possible.

Case 2: Cardinality = 2

Assume S = {n, m} where n is the smallest.

Since n cannot be a perfect square, the first step is n → n + 3.

For a loop of two elements, the sequence must be n → n + 3 → n. This means n + 3 must be a perfect square, and √(n + 3) = n.

Solving n + 3 = n² gives n² - n - 3 = 0. Using the quadratic formula, n = (1 ± √13) / 2, which is not an integer.

Conclusion: A trajectory of cardinality 2 is not possible.

Sub-Problem B: Show That {3, 6, 9} Is The Only Trajectory Of Cardinality Three.

Assume a trajectory S with |S| = 3. Let n be the smallest element. As n cannot be a perfect square, the sequence starts n → n + 3.

For a three-element loop n → n + 3 → n + 6 → n, n + 3 must not be a perfect square (e.g., if n=3, n+3=6 is not a perfect square).

This implies n + 6 must be a perfect square, and √(n + 6) = n.

Solving n + 6 = n² gives n² - n - 6 = 0. Factoring yields (n - 3)(n + 2) = 0.

Since n must be a positive integer (a > 1), we get n = 3.

Conclusion: The only possible trajectory of cardinality three is {3, 6, 9}.

Sub-Problem C: Show That For Any Integer k ≥ 3, There Is A Trajectory Of Cardinality k.

• For k = 3: We know {3, 6, 9} is a trajectory of cardinality 3.

• For k > 3: We can construct larger trajectories by prefixing the 3-6-9 loop with perfect squares.

• Example: Trajectory of Cardinality 4
  Start with 36: 36 → √36 = 6 → 6 + 3 = 9 → √9 = 3 → 3 + 3 = 6 (enters loop).
  Trajectory: {36, 6, 9, 3}. This has cardinality 4.

• Example: Trajectory of Cardinality 5
  Start with 1296 (36²): 1296 → √1296 = 36 → √36 = 6 → 6 + 3 = 9 → √9 = 3 → 3 + 3 = 6 (enters loop).
  Trajectory: {1296, 36, 6, 9, 3}. This has cardinality 5.
  By repeatedly taking squares of numbers that eventually lead to the 3-6-9 loop, we can create initial chains of any desired length, thus constructing trajectories of any cardinality k ≥ 3.

Sub-Problem D: Find An Infinite Trajectory.

An infinite trajectory exists if the sequence never reaches a perfect square and thus never loops or terminates.

Key Mathematical Concept: Properties of Numbers Modulo 3

1. Squares Modulo 3: Any integer x² is congruent to either 0 or 1 modulo 3.

• x ≡ 0 (mod 3) ⇒ x² ≡ 0 (mod 3)

• x ≡ 1 (mod 3) ⇒ x² ≡ 1 (mod 3)

• x ≡ 2 (mod 3) ⇒ x² ≡ 1 (mod 3)
  Implication: A number congruent to 2 (mod 3) cannot be a perfect square.

2. Game Rule with Modulo 3: Adding 3 does not change the number's congruence modulo 3.

• a + 3 ≡ a (mod 3)

Constructing an Infinite Trajectory:

Consider starting with any a > 1 such that a ≡ 2 (mod 3).

• Since a ≡ 2 (mod 3), a is not a perfect square, so a → a + 3.

• a + 3 will also be ≡ 2 (mod 3), so it's not a perfect square, and the rule + 3 applies again.
  This pattern a, a+3, a+6, ... continues indefinitely. All numbers remain ≡ 2 (mod 3) and none will ever be a perfect square. Thus, the √a rule is never applied, creating an infinite trajectory.
  Example: Starting with a = 5 (since 5 ≡ 2 (mod 3)) yields 5, 8, 11, 14, ..., an infinite trajectory.

Solution To Main Problem: Which Numbers Have A Finite Trajectory?

Based on the analysis:

• Numbers where a > 1 and a ≡ 2 (mod 3) generate infinite trajectories.

• Therefore, finite trajectories must come from numbers a > 1 where a ≡ 0 (mod 3) or a ≡ 1 (mod 3).

For a trajectory to be finite, the sequence must eventually lead to the 3-6-9 cycle. This means the sequence must never generate a number x such that x ≡ 2 (mod 3) and x is not a perfect square. If a number X in the sequence is a perfect square, its square root √X is taken. If √X ≡ 2 (mod 3), the trajectory becomes infinite (e.g., 16 → 4 → 2, and 2 ≡ 2 (mod 3) leads to an infinite sequence).

The final condition for a finite trajectory is that all numbers generated within the sequence are either ≡ 0 (mod 3) or ≡ 1 (mod 3) OR if a number x ≡ 2 (mod 3) is generated, it immediately leads to a perfect square P such that √P (and all subsequent numbers in its trajectory) are themselves part of a finite trajectory. This ultimately means the trajectory must always lead to the 3-6-9 loop.

CMI 2024 B1 - Finding Odd Integers n > 1

Problem Statement: Determine all odd integers n > 1 such that n is a factor of 2023^n - 1.

This is equivalent to 2023^n ≡ 1 (mod n).

Part 1: Find The Two Smallest Such Integers.

We test odd integers n > 1:

• Test n = 3:
  2023 ≡ 1 (mod 3) (since 2023 = 3 * 674 + 1).
  So, 2023^3 ≡ 1^3 ≡ 1 (mod 3).
  Conclusion: n = 3 satisfies the condition. This is the first smallest integer.

• Test n = 5:
  2023 ≡ 3 (mod 5) (since 2023 = 5 * 404 + 3).
  2023^5 ≡ 3^5 = 243 ≡ 3 (mod 5).
  Since 3 ≠ 1 (mod 5), n = 5 does not satisfy the condition.

• Test n = 7:
  2023 ≡ 0 (mod 7) (since 2023 = 7 * 289 + 0).
  2023^7 ≡ 0^7 ≡ 0 (mod 7).
  Since 0 ≠ 1 (mod 7), n = 7 does not satisfy the condition.

• Test n = 9:
  2023 ≡ 7 (mod 9) (since 2023 = 9 * 224 + 7).
  We need to check 2023^9 ≡ 1 (mod 9).
  2023^9 ≡ 7^9 (mod 9). Since 7 ≡ -2 (mod 9),
  7^9 ≡ (-2)^9 = -512 (mod 9).
  -512 = -57 * 9 + 1, so -512 ≡ 1 (mod 9).
  Conclusion: n = 9 satisfies the condition. This is the second smallest integer.

The two smallest integers satisfying the condition are 3 and 9.

Part 2: Prove That There Are Infinitely Many Such Integers.

Claim: If k is a solution, then 3k is also a solution. (i.e., if k | (2023^k - 1), then 3k | (2023^(3k) - 1)).

Proof by Induction:

1. Base Case: We know k = 3 is a solution. The claim suggests 3 * 3 = 9 should also be a solution, which we confirmed.

2. Inductive Step:
   Assume 2023^k ≡ 1 (mod k). We need to show 2023^(3k) ≡ 1 (mod 3k).
   Consider 2023^(3k) - 1. Using the difference of cubes formula A^3 - B^3 = (A - B)(A^2 + AB + B^2):
   2023^(3k) - 1 = (2023^k)^3 - 1^3 = (2023^k - 1) * (2023^(2k) + 2023^k + 1).

• From our assumption, (2023^k - 1) is divisible by k.

• Now, we show (2023^(2k) + 2023^k + 1) is divisible by 3.
  We know 2023 ≡ 1 (mod 3).
  So, 2023^(2k) ≡ 1^(2k) ≡ 1 (mod 3).
  And 2023^k ≡ 1^k ≡ 1 (mod 3).
  Therefore, 2023^(2k) + 2023^k + 1 ≡ 1 + 1 + 1 ≡ 3 ≡ 0 (mod 3).
  This proves the second factor is divisible by 3.

Since one factor is divisible by k and the other by 3, their product (2023^(3k) - 1) is divisible by 3k.
Starting with n = 3, this proves that 3^m for m ≥ 1 (i.e., 3, 9, 27, 81, …) are all solutions. Thus, there are infinitely many such integers.

CMI 2023 B6 - Perfect Squares From Factorials

Problem Statement: Suppose n > 1 is a natural number which is not congruent to 3 (mod 4). Prove that there exist integers i and j such that 1 ≤ i ≤ j ≤ n, and the expression P = (1! * 2! * 3! * ... * n!) / (i! * j!) is a perfect square.

Constraints on n: n can be 0 (mod 4), 1 (mod 4), or 2 (mod 4).

Key Observation for Perfect Squares: m! * (m-1)! = m * ((m-1)!)^2.

Case 1: n ≡ 0 (mod 4)

Let n = 4k for some positive integer k.

1. ttNumerator (N): N = 1! * 2! * ... * (4k)!.
   Group adjacent factorials:
   N = [2! * 1!] * [4! * 3!] * ... * [(4k)! * (4k-1)!]
   N = [2 * (1!)^2] * [4 * (3!)^2] * ... * [4k * ((4k-1)!)^2]
   N = [ ( (4k-1)! )^2 * ... * (3!)^2 * (1!)^2 ] * [ 4k * (4k-2) * ... * 4 * 2 ]
   The term [ ( (4k-1)! )^2 * ... * (3!)^2 * (1!)^2 ] is a perfect square, let it be m².

2. Remaining Product (P_rem): P_rem = 4k * (4k-2) * ... * 4 * 2.
   This is a product of 2k even numbers. Factor out 2 from each:
   P_rem = 2^(2k) * [ 2k * (2k-1) * ... * 2 * 1 ] = 2^(2k) * (2k)!
   Since 2^(2k) is a perfect square (2^k)², we can write N = m² * (2^k)² * (2k)! = M² * (2k)! where M = m * 2^k.

3. Finding i and j:
   We need P = (M² * (2k)!) / (i! * j!) to be a perfect square.
   If we choose i = 1 and j = 2k:
   1 ≤ 1 ≤ 2k ≤ 4k = n (since k ≥ 1). These choices are valid.
   Then P = (M² * (2k)!) / (1! * (2k)!) = M² / 1 = M².
   Since M² is a perfect square, we have found a valid pair (i, j) = (1, 2k) for n ≡ 0 (mod 4). This proves the existence.
   (Similar proofs exist for n ≡ 1 (mod 4) and n ≡ 2 (mod 4).)

Putnam 2004 - Determining All Positive Integers n

Problem Statement: Determine all positive integers n for which there exist positive integers a, b, and c satisfying the equation:

2a^n + 3b^n = 4c^n

1. Initial Analysis And Base Case (n = 1)

• Consider n = 1: The equation becomes 2a + 3b = 4c.
  The right side 4c is always even, so 2a + 3b must be even. Since 2a is even, 3b must be even, implying b must be an even integer.
  Let's find a solution:
  Set b = 2. Equation: 2a + 6 = 4c ⇒ a + 3 = 2c.
  For a + 3 to be even, a must be an odd integer.
  Set a = 1. Equation: 1 + 3 = 2c ⇒ 4 = 2c ⇒ c = 2.
  Thus, for n = 1, (a, b, c) = (1, 2, 2) is a valid solution with positive integers.
  Conclusion: n = 1 is a possible value.

2. Divisibility By 2 Analysis (General n)

To simplify, we look for primitive solutions, where gcd(a, b, c) = 1. If a solution (a,b,c) exists, a primitive solution (x,y,z) also exists by dividing by d = gcd(a,b,c).

Equation: 2a^n + 3b^n = 4c^n

• 2a^n is divisible by 2.

• 4c^n is divisible by 2.

• By the divisibility rule (if all terms but one in a sum are divisible by a prime, the remaining term must also be), 3b^n must be divisible by 2.
  Since 3 is odd, b^n must be divisible by 2, which means b must be divisible by 2.
  Let b = 2b1. Substitute: 2a^n + 3(2b1)^n = 4c^n
  2a^n + 3 * 2^n * b1^n = 4c^n
  Divide by 2: a^n + 3 * 2^(n-1) * b1^n = 2c^n.

• Now, 3 * 2^(n-1) * b1^n is divisible by 2 (for n ≥ 1), and 2c^n is divisible by 2.
  Therefore, a^n must be divisible by 2, which means a must be divisible by 2.
  Let a = 2a1. Substitute: (2a1)^n + 3 * 2^(n-1) * b1^n = 2c^n
  2^n * a1^n + 3 * 2^(n-1) * b1^n = 2c^n
  Divide by 2: 2^(n-1) * a1^n + 3 * 2^(n-2) * b1^n = c^n.

3. Establishing An Upper Bound For n

We found that a is even and b is even.

If c were also even, then gcd(a, b, c) would be at least 2, contradicting our assumption of a primitive solution (gcd(a, b, c) = 1).

Therefore, c must be an odd integer.

Consider the equation 2^(n-1) * a1^n + 3 * 2^(n-2) * b1^n = c^n.

Since c is odd, c^n must also be odd.

For the left side to be odd, exactly one of the terms 2^(n-1) * a1^n or 3 * 2^(n-2) * b1^n must be odd.

• If n-1 = 0, then n = 1.
  The equation becomes a1^1 + 3 * 2^(-1) * b1^1 = c^1, which involves a fraction, so n-1 cannot be 0 if n-2 is negative.

• If n-2 = 0, then n = 2.
  The equation becomes 2^(2-1) * a1^2 + 3 * 2^(2-2) * b1^2 = c^2
  2a1^2 + 3 * 1 * b1^2 = c^2
  2a1^2 + 3b1^2 = c^2.
  In this case, 2a1^2 is even. For c^2 to be odd, 3b1^2 must be odd, implying b1 is odd.
  (This is a valid scenario for n=2).

• If n > 2:
  Then n-1 ≥ 2 and n-2 ≥ 1.
  This means 2^(n-1) * a1^n is even, and 3 * 2^(n-2) * b1^n is even.
  The sum of two even numbers is always even. So, if n > 2, then c^n would be even, implying c is even. This contradicts our finding that c must be odd for primitive solutions.

Conclusion: n cannot be greater than 2.

4. Analysis For n = 2

We must investigate n = 2. The equation becomes: 2a² + 3b² = 4c².

• We know a and b are even, and c is odd.

• Rearrange: 3b² = 4c² - 2a².
  Adding a² to both sides: 3a² + 3b² = a² + 4c²
  3(a² + b²) = a² + 4c².
  Since the left side is divisible by 3, the right side a² + 4c² must be divisible by 3.
  This means a² + c² + 3c² is divisible by 3, so a² + c² must be divisible by 3.

• Modular Properties of Squares (Mod 3): (Remember: Any integer square is congruent to 0 or 1 modulo 3.)
  For a² + c² ≡ 0 (mod 3), the only possibility is that both a² ≡ 0 (mod 3) and c² ≡ 0 (mod 3).
  This implies a is divisible by 3 and c is divisible by 3.

• Contradiction with gcd(a, b, c) = 1:
  Since a and c are divisible by 3, let a = 3a1 and c = 3c1.
  Substitute these into 2a² + 3b² = 4c²:
  2(3a1)² + 3b² = 4(3c1)²
  18a1² + 3b² = 36c1²
  Divide by 3: 6a1² + b² = 12c1².
  Now, 6a1² is divisible by 3, and 12c1² is divisible by 3.
  Therefore, b² must be divisible by 3, which implies b is divisible by 3.

We have concluded that for n = 2, a, b, and c are all divisible by 3. This means gcd(a, b, c) is at least 3, contradicting our assumption of a primitive solution (gcd(a, b, c) = 1).

Therefore, no primitive integer solution exists for n = 2.

5. Final Conclusion

Since n cannot be greater than 2, and n=2 has no solutions (based on primitive solution analysis), the only positive integer n for which solutions (a, b, c) exist for 2a^n + 3b^n = 4c^n is n = 1.