What separates a Level 5 problem from a Level 7, side by side

Students who ask what separates a Level 5 problem from a Level 7 usually expect the answer to be harder. It mostly is not. Two of the real IMO proofs on Lemma make the point better than any definition could. IMO 1964, Problem 2 and IMO 1995, Problem 2 are both inequalities in three variables, both rated 5 out of 6 for difficulty on their problem pages, and both expected to take under an hour. The ladder puts the first at Level 5 and the second at Level 7.

So the gap between the two levels is not effort or stamina. It is two specific things, and both are visible once the problems sit next to each other.

What the two levels are on the ladder

The levels are not difficulty bands invented for this page. They are the rungs Lemma teaches in order, and each technique page says where it unlocks.

• Level 5, Olympiad Initiate: "the proof-writing gateway". Induction, invariants and monovariants, without loss of generality, Diophantine equations, the power of a point, Ceva and Menelaus, double counting, symmetric sums.
• Level 6, Olympiad Veteran: games, functional equations, graphs, quadratic residues, Vieta jumping, the incircle configuration.
• Level 7, IMO Candidate: "the theorems that power shortlist problems". Orders and Fermat's little theorem, CRT and Euler, Wilson, Pell's equation, Cauchy–Schwarz beyond the AM–GM first steps of Level 4, convexity and Jensen, Burnside, directed angles, barycentric coordinates, and a lesson on managing a whole olympiad paper.

Read the two lists and the first difference is already there. Level 5 is a set of methods: ways of organising a proof that work on almost any topic. Level 7 is a set of theorems: specific results that are each the key to a particular family of problems, and useless outside it.

The inequality pair

Level 5. IMO 1964/2: if $a, b, c$ are the sides of a triangle, then $$a^2(b+c-a) + b^2(c+a-b) + c^2(a+b-c) \le 3abc.$$ Move everything to one side and expand. The difference $3abc - \text{LHS}$ is exactly $$a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b),$$ which is Schur's expression. It is symmetric, so you may assume $a \ge b \ge c$. Then the first two terms combine to $(a-b)^2(a+b-c)$, which is at least zero, and the last term is a product of two non-positive factors with $c \ge 0$, so it is at least zero too. Done, with equality only for the equilateral triangle.

Every move there is a Level 5 move: expand, notice symmetry, order the variables, check signs. The problem never asks you to see it as anything other than what it is. It even turns out the triangle condition is not needed: the inequality holds for any non-negative $a, b, c$, which is the kind of thing you find by doing the algebra honestly.

Level 7. IMO 1995/2: if $a, b, c > 0$ and $abc = 1$, then $$\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \ge \frac{3}{2}.$$ Expand this and you drown. The problem does not yield until you change what it looks like. Put $x = 1/a$, $y = 1/b$, $z = 1/c$; then $xyz = 1$ still, and each term collapses to $\frac{x^2}{y+z}$. Only now does a named theorem apply: Cauchy–Schwarz in Engel form gives a sum of at least $\frac{x+y+z}{2}$, and AM–GM with $xyz = 1$ gives $x + y + z \ge 3$. Equality at $a = b = c = 1$, where the sum is exactly $\frac32$.

That is the pattern. The Level 5 problem is solved by working on it as given. The Level 7 problem is solved by a chain of three moves, and the first one is a transformation the statement gives no hint of. Nothing in "$abc = 1$" says take reciprocals. You have to have seen that a constraint like that is often an invitation to substitute.

The number theory pair

The same split shows up in number theory.

Level 5. Prove that $x^2 + y^2 = 3z^2$ has no solutions in positive integers. The 3 in the equation tells you where to look. Squares are 0 or 1 modulo 3, so $x^2 + y^2$ is divisible by 3 only when both $x$ and $y$ are, and then a descent finishes it. Modular arithmetic is the first tool you would try, and the problem rewards the first try.

Level 7. IMO 1986, Problem 1: if $d$ is a positive integer other than 2, 5 and 13, show that some two distinct numbers from $\{2, 5, 13, d\}$ have $ab - 1$ not a perfect square. The modulus is not in the statement, and you have to earn it. First notice that the three pairs without $d$ all do give squares: $9$, $25$ and $64$. So the claim is really that $2d - 1$, $5d - 1$ and $13d - 1$ cannot all be squares. If $2d - 1$ is an odd square, $d$ is odd. Then $5d - 1$ and $13d - 1$ are even squares, their difference $8d$ forces a difference of two squares equal to $2d$, and a difference of squares is never 2 modulo 4.

None of those steps uses a theorem beyond Level 4. What makes it Level 7 is the setup: check the easy cases, negate the claim correctly (all three are squares, not one), extract the parity of $d$, and only then find the congruence that bites. Four moves, and the problem hands you none of them.

So what actually separates them

Put the pairs together and three differences come out, in order of how often they decide the problem.

1. The first move is hidden. A Level 5 problem tells you what kind of problem it is: "for every $n$" points at induction, a process with moves points at an invariant, a 3 in the equation points at mod 3. A Level 7 problem usually needs a substitution or a reformulation before it looks like anything.
2. The chain is longer. Level 5 proofs tend to be one idea carried through. Level 7 proofs are two or three ideas in sequence, and losing the thread in the middle is the most common way to lose the marks.
3. The theorem is specific. Cauchy–Schwarz, orders, Jensen, CRT: each is the right tool for a narrow family and the wrong tool everywhere else. Knowing the statement is not enough; you need to recognise its shape after a transformation.

How to cross the gap

Do not jump to Level 7 problems and hope. The methods at Level 5 are what the Level 7 chains are made of: the reciprocal substitution only helps because you can then expand, order and sign-check without thinking.

What helps is practising the hidden first move on purpose. When you finish a Level 7 problem, write down only the first line, the substitution or reformulation, and ask what in the statement should have suggested it. "$abc = 1$ with cubes in the denominators" is a signal you can learn; "the modulus is not given, so find a parity fact first" is another. After a dozen of those, the first move stops being hidden.

If you are not sure which rung you are on, the free Find Your Level quiz places you in a few minutes. The problem archive carries every statement above in full with its difficulty and expected time, and the technique index shows where each method unlocks on the ladder.