INMO — Indian National Mathematical Olympiad — Answer Writing
Olympiad marking is close to binary on rigour: a brilliant idea with a gap in the argument scores far below a plodding proof that closes every case. Writing the proof is the skill, and it is separable from solving the problem.
Attempt all problems. Justify every step; a correct answer without a complete argument earns no credit. Calculators are not permitted.
- 1.Find all positive integers n such that n² + 1 divides n³ + 3.17
- 2.Let ABC be a triangle with circumcircle ω. Prove that the reflection of the orthocentre in BC lies on ω.17
- 3.Determine all functions f : ℝ → ℝ satisfying f(x + f(y)) = f(x) + y for all real x, y.17
One INMO answer, from question to marked report
A real question, an attempt with the weaknesses a real attempt has, and the report an AI examiner hands back — in the marking unit INMO genuinely uses. This is the whole product; there is nothing else to sign up for.
Scoped to the paper and topic you pick, at the marks it is genuinely set at.
Find all positive integers n such that n squared plus 1 divides n cubed plus 3.
Type it, dictate it, or write on paper and photograph the page — your handwriting is read.
Photographed, read, and marked in red — every expected step scored down the right margin and the omissions written in.
We have n^3 + 3 = n(n^2 + 1) - n + 3. So n^2 + 1 divides -n + 3, that is n^2 + 1 divides n - 3. For large n, n^2 + 1 is bigger than n - 3, so we need n - 3 = 0 or small n.? Checking, n = 3 works and n = 1 works. So the answers are n = 1 and n = 3. First note that n^2 + 1 is positive for every positive integer n, so the divisibility statement makes sense throughout. Writing the division the other way, n^3 + 3 = n(n^2 + 1) + (3 - n), which is the same relation with the sign changed, and a divisor of a number also divides its negative, so it makes no difference whether we work with 3 - n or with n - 3. Since n^2 + 1 grows quadratically while n - 3 grows linearly, the divisibility can only hold when the linear part is zero or is small in comparison. Trying small values, n = 1 gives 2 dividing 4 which is true, and n = 3 gives 10 dividing 30 which is also true.
- The explicit inequality and the value of n from which it holds
- Checking every remaining case, n equals 1, 2 and 3, visibly
- A closing sentence asserting that these are all solutions
Model answer for this question
Reduce: n cubed plus 3 equals n times (n squared plus 1) minus n plus 3, so n squared plus 1 divides n minus 3. Bound: if n is at least 4 then n squared plus 1 exceeds the absolute value of n minus 3, forcing n minus 3 equals 0, giving n equals 3. Check the remaining cases n equals 1, 2, 3 directly. Conclude: the complete solution set is n in {1, 3}, and state that no other n can occur by the bound.
Photograph the page from your phone and the Studio reads your handwriting — the only way to practise the speed and layout the paper actually tests.
Fast enough to write, read the report and rewrite the same answer in one sitting — which is where the improvement actually comes from.
The scoring behind those pen marks: every expected step, what it was worth, what you earned, and a model answer.
- Reduction to a divisibility of lower degree5/5
Correct and complete.
- Size argument stated as an inequality2/5
Asserted informally. Write: for n greater than or equal to 4, n squared plus 1 is greater than the absolute value of n minus 3, so the quotient must be zero.
- Exhaustive check of the remaining finite cases2/4
n equals 1 and 3 are found; n equals 2 is never checked and the reader cannot see that the list is complete.
- Conclusion stated as a complete solution set2/3
The set is stated but not asserted as exhaustive.
INMO at a glance
| Conducting body | Homi Bhabha Centre for Science Education (HBCSE) |
|---|---|
| Mode | Handwritten proofs |
| Papers | 4 |
| Total | Six problems over four and a half hours |
| Marking | INMO — step marking |
What INMO sets
The papers you sit, and what each is worth.
Functional equations, polynomials, inequalities.
Counting, extremal arguments, invariants.
Synthetic and computational, with full justification.
Divisibility, congruences, Diophantine equations.
- Marked to the INMO scheme
- Model answer, every time
- Handwriting read from a photo
- AI viva on any topic
- Full timed papers
- Every attempt kept in your archive
- WriteA question, answered and marked.
- VivaQuestioned on Functional equations and more.
- Full paperA timed paper, one scorecard.