IOQM common mistakes
IOQM · Understand the traps and how to avoid them.
On a diophantine or divisibility problem you test a handful of small values, find one solution, and write it as the answer without proving it is the only one or addressing the full range of the variable.
IOQM problems often have a unique answer hidden behind a general condition; small-case testing feels like verification, and the psychological relief of having 'an answer' stops further search. Candidates also confuse 'a solution exists' with 'find all solutions'.
What to do instead
Adopt a written protocol: after finding a candidate, explicitly answer three questions in the margin - (1) does the problem ask for all values or one value, (2) can I construct a second solution by varying a parameter, (3) which modular invariant or size argument rules out others. Practise with problems where the small case is a decoy (e.g. x^2 = 2^y + 1 style), and always finish with a one-line uniqueness argument or a note that none is needed.
You apply a theorem without checking its hypotheses - using Fermat's little theorem with a composite modulus, dividing both sides of a congruence by a non-coprime number, or cancelling a factor in a modular equation.
Theorems are memorised as formulas and the hypothesis is treated as boilerplate. Under time pressure the mind pattern-matches 'prime-looking modulus' or 'cancel common factor' from ordinary algebra, where cancellation is always legal.
What to do instead
Keep a one-page 'licence to use' card: FLT needs p prime and gcd(a,p)=1; Euler needs gcd(a,n)=1; cancellation mod n needs gcd(factor,n)=1. Before writing any congruence step, say the hypothesis aloud or write 'gcd check' next to it. Drill ten problems specifically designed to break these hypotheses (mod 8, mod 12, non-coprime factors) so the check becomes automatic.
On a functional equation you substitute one or two convenient values (x=0, x=1, y=x) and then announce the function, without verifying that the candidate satisfies the equation for all reals or checking whether other families fit the same data.
Substitution is the standard first move and gives quick partial information; the brain treats partial information as complete. Candidates also skip the final verification step because it feels redundant after 'solving'.
What to do instead
Treat every FE as a three-stage task written out in order: Stage 1 collect values (P(0), P(1), P(x,x), P(x,0)); Stage 2 derive regularity (injectivity, surjectivity, monotonicity, or boundedness on an interval); Stage 3 verify the candidate by substitution into the original equation and argue uniqueness from the regularity proved. Never write the answer before Stage 3 is on paper. Practise at least five FEs where a spurious solution (e.g. f=0 alongside f(x)=x) survives Stage 1.
You manipulate inequalities by squaring, cross-multiplying, or applying AM-GM without recording the sign conditions, equality conditions, or domain restrictions, and then present the bound as if equality were attainable.
Algebraic habit treats inequalities like equations. AM-GM especially invites blind application; the equality condition is checked only when the problem explicitly asks for it, but IOQM answer-type problems need the extremal value, which requires attainability.
What to do instead
For every inequality write three lines before computing: signs of all quantities, the domain, and the equality condition of each named inequality. After obtaining a bound, exhibit the explicit tuple achieving it, or state why it cannot be achieved and tighten. Build a habit of testing the equality case first with a numerical example to see whether the bound is sharp.
You treat a polynomial or sequence problem as if roots, coefficients, and indices were interchangeable - assuming integer roots when only real roots are given, or summing a series from index 1 when it starts at 0.
Vieta and AP/GP formulas are recalled as symbol patterns, and index conventions differ between textbooks. Candidates also import 'integer' from the answer format (IOQM answers are integers) into the problem data, which is a category error.
What to do instead
Before solving, copy the given conditions verbatim into a 'given / wanted / allowed' box, explicitly noting the number system, the index range, and whether roots are distinct. When using Vieta, write the polynomial in the exact stated form first. For series, write the first and last terms explicitly with their indices before applying any formula.
You draw a diagram, spot an apparent concurrency, collinearity, or right angle, and use it as a fact in the proof because it looks true in your figure.
Geometry is visual, and accurate sketches are persuasive. The default configuration in a hand drawing (acute triangle, points inside) hides the cases where the claimed property fails, and the brain does not distinguish 'observed' from 'proved'.
What to do instead
Maintain a two-column discipline: left column 'proved', right column 'conjectured from figure'. A conjectured fact may guide construction but cannot enter a chain of reasoning until proved. Redraw every configuration in a deliberately awkward form (obtuse, points on extensions, crossing chords) and check whether the conjecture survives; if it does not, the original claim needs extra hypotheses.
You prove a special case of a geometry statement - for instance assuming the triangle is isosceles, or placing the origin at a convenient vertex - and present it as the general proof.
Synthetic proofs often become easy after an unstated symmetry assumption, and coordinate geometry invites choosing coordinates that silently impose a condition (e.g. setting a side to 1 when the result is not scale-invariant, or putting a vertex at the origin when the result depends on its position).
What to do instead
After finishing a proof, list every assumption used and mark which were given. For coordinate solutions, verify that the coordinate choice uses only allowed freedoms (translation, rotation, scaling) and that the final expression is homogeneous or invariant as required. For synthetic solutions, ask 'which step fails if the triangle is scalene?' and repair it before moving on.
You count arrangements by multiplying choices without deciding whether order matters and whether objects are distinct, producing a total that is off by a factorial or a symmetry factor.
Permutation and combination formulas look similar and are selected by keyword ('arrange' vs 'choose'), but real problems mix the two - arranging identical objects, circular seating with reflections, or selecting then arranging. Overcounting by symmetry is invisible in the final number.
What to do instead
Before computing, write a one-line model: 'I am counting [functions / subsets / necklaces] from a set of size X to size Y', which fixes whether order and repetition are allowed. For any answer involving symmetry (circles, bracelets, grids with rotations), divide by the group order explicitly and check with a small case (n=3 or n=4) by hand-listing all objects.
You apply the pigeonhole principle or an extremal argument with a vague choice of 'boxes' or of the extremal object, then assert the conclusion without showing the boxes are well defined or that the extremal object has the needed property.
Pigeonhole is stated as a slogan and the difficulty lies entirely in constructing the partition, which candidates skip because the conclusion feels immediate once boxes are named. Extremal arguments similarly need a proof that the chosen maximum exists and satisfies the constraints.
What to do instead
Write the pigeonhole argument in a fixed template: define the set of pigeons, define the map to holes, count both, state the inequality, then conclude. If the boxes are not obvious, work backwards from the desired conclusion ('what would force two objects to share a property?'). For extremal arguments, verify the maximum exists (finite set) and that perturbing it is impossible, since that impossibility is the actual proof.
You spend the first twenty minutes on a single hard problem, then rush the remaining ones and lose easy marks to arithmetic slips and unfinished reasoning.
IOQM rewards breadth; a single elegant solution is worth the same as several short ones, but the sunk-cost effect keeps candidates on a problem once invested. Without a timing plan, the difficulty ordering is discovered too late.
What to do instead
Adopt a three-pass strategy with a watch: Pass 1 (about 40 percent of time) attempt every problem briefly and record a confidence rating; Pass 2 solve all high-confidence problems completely; Pass 3 return to the rest. Enforce a hard cutoff (e.g. eight minutes) on any single problem in Pass 1. Rehearse this on full-length timed mock papers so the switching behaviour is practised, not just intended.