Proof by cases splits the domain into pieces that together cover everything, then proves the claim on each piece. Parity is the workhorse: every integer is even (n = 2k) or odd (n = 2k + 1), and that single split settles n² + n even, the parity of n² matching n, and remainders of n² on division by 4 being 0 or 1. On each branch, substitute the form, expand, and factor out the claimed divisor explicitly — the algebra is the proof.
Remainder splits generalise parity: non-multiples of 3 are exactly 3k + 1 or 3k + 2, so squaring both forms shows n² leaves remainder 1; cubes reduce mod 9 the same way, and the division algorithm covers negatives and zero too. Absolute values demand kink cases instead: |x| splits at 0 (x ≥ 0 versus x < 0), |x − 1| + |x + 1| needs all three intervals around ±1, and |xy| = |x||y| needs the sign quadrants resolved one by one.
The fatal error is the missing case: 'n even, so n(n + 1) is even' abandons half the integers — for odd n it is the factor n + 1 = 2k + 2 that supplies the factor of 2. Dropping x = 0 from an absolute-value split, or covering only two of three kink intervals, fails the same way. Method: name the classification principle first (parity, remainder, sign, kink), verify it is exhaustive before writing algebra, and remember overlap is wasteful but harmless while omission is fatal. In the drill set, every cases question is tagged with the split it expects, so a miss tells you precisely which classification principle to rehearse — parity, remainder, sign, or kink.
Study method: attempt five timed questions, log every miss with its topic code, redo misses from scratch within 48 hours, then re-attempt the topic a week later. Pair each hub with its drill link below and a fortnightly full mock.
Place this topic in the wider test with the syllabus checklist and Paper 1 vs Paper 2, learn the timing system in how to prepare, and check what scores mean in scores explained.