How to get better at proofs when you can only compute

There is a specific moment where a strong computational student stalls. You can factor, substitute, chase an algebraic identity, and land a hard multiple-choice question in three minutes. Then the word prove appears at the front of a problem and the page stays blank, not because you cannot do the arithmetic, but because you do not know what would count as finishing.

The usual diagnosis is that proof is a different talent, arriving later or not at all. That is wrong, and believing it costs people years. A proof is a computation with its quantifier made explicit. You already do the hard half. What is missing is a small set of moves that convert I checked it and it worked into it cannot fail, and each of those moves is itself a computation.

Move one: find out whether the case list is finite

Sometimes checking every case is a complete proof, and people who can only compute walk past their own best method because it feels like cheating.

Take prove that $n^5 - n$ is divisible by 30 for every integer $n$. There are infinitely many integers, so brute force looks hopeless. It is not. Divisibility by 30 depends only on $n$ modulo 30, because replacing $n$ by $n + 30$ changes $n^5 - n$ by a multiple of 30. So the infinitely many cases collapse to thirty, and thirty cases is a finite computation you can do by hand.

That is a real proof, written in the language you already speak. The skill being trained is not arithmetic, it is the question that comes first: is there a reason this problem has only finitely many genuinely different cases? Periodicity, symmetry, and a bounded parameter are the three usual reasons.

Move two: know why forty checks prove nothing

The counterweight matters more than the move. Consider $n^2 + n + 41$. It is prime at $n = 0$, at $n = 1$, and at every value up to $n = 39$. Forty consecutive successes is more evidence than most people would ever demand. At $n = 40$ it equals 1681, which is $41^2$, and the pattern dies.

The gap between move one and move two is the entire difference between computing and proving, and it is one sentence wide.

Move three: when the list is infinite, compute the step

If the cases cannot be collapsed, stop trying to compute the statement and compute the difference between neighbouring cases instead. That is all induction is, and phrased that way it is a computational technique rather than a formal ritual.

Prove that a set with $n$ elements has exactly $2^n$ subsets is the cleanest example. Counting subsets directly gets unpleasant fast. Counting the change is trivial: every subset of an $n$-element set either contains the new element or does not, and those two families are the same size, so the count doubles. One line of arithmetic, and it covers every $n$ at once.

Notice what you actually computed. Not a number, a ratio between consecutive cases. When an induction feels impossible, it is almost always because the quantity you chose does not have a clean relationship to its predecessor. Change the quantity, not the method.

Move four: when nothing can be sampled, compute an identity

Some statements have no case list at all. Prove that $x^4 - x + \tfrac12 > 0$ for all real $x$ is one: the reals are not enumerable, and sampling is genuinely dangerous here because the polynomial dips close to zero. Its minimum is about 0.0275, near $x \approx 0.63$, so a coarse table of values could easily miss how tight it gets.

The proof is a pure computation, just an algebraic one. Multiply by 4 and rearrange:

$$4x^4 - 4x + 2 = (2x^2 - 1)^2 + (2x - 1)^2$$

Expand the right side and you get the left, which anyone who can compute can verify in a minute. A sum of two squares is at least zero, and the two squares cannot vanish at the same time, since $2x - 1 = 0$ forces $x = \tfrac12$, where $2x^2 - 1 = -\tfrac12$. So the total is strictly positive everywhere.

This is the move that converts most inequality problems. You are not looking for insight, you are looking for a rewriting that makes the conclusion visible, and rewriting is something computational students are better at than they think.

Move five: when the object is unknown, compute the extreme one

The hardest proofs to start are the ones about some object you have never been shown: some triangle, some configuration, some arrangement. The extremal principle fixes that by making you choose one, namely the largest or smallest, and the choice hands you an inequality for free.

Among all triangles with a given perimeter, the equilateral one has the largest area is the model. You cannot check every triangle. You can take the one with the largest area and ask what must be true of it. If two of its sides differed, you could nudge them towards each other, keep the perimeter fixed and increase the area, contradicting the choice. So no two sides differ.

The computation there is the nudge, and it is ordinary algebra. The proof structure was supplied entirely by the decision to look at the extreme case.

The workflow that puts them together

None of this says to stop computing. The honest order of work runs the other way round.

1. Compute first, to find the answer. In the locker problem, a hundred students toggling a hundred lockers, simulate it. The lockers left open are 1, 4, 9, 16 and so on up to 100, ten of them, all perfect squares. Now you know what to prove.
2. Ask why the answer is what it is. A locker ends open when it was toggled an odd number of times, and it is toggled once per divisor. Divisors pair up as $d$ and $n/d$, so the count is odd only when a divisor pairs with itself, which happens exactly for perfect squares.
3. Write the sentence that covers every case. One line, naming the quantifier. This is the part that gets skipped and the only part a marker cares about.

Steps 1 and 2 are computation. Step 3 is what you are actually short of, and it is much smaller than it looks from the outside.

What to practise this week

Do not start with proof-only problems. Take problems you can already solve numerically and write the missing sentence for each: this holds in every case, because …. Ten of those, done properly, move you further than one heroic attempt at an olympiad question.

Then go looking for the two moves you do not have yet. Most computational students already own moves one, two and four, and are missing three and five entirely. The technique pages name each one and say where it sits in the ladder, the problem archive has the statements above in full with difficulty and expected time, and the daily problem posts a fresh one in three tiers with no account needed, which is enough to run the write-the-sentence drill every morning. If you want to know which moves you are missing rather than guess, the diagnostic is the shortest route to an answer.