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TMUA 2017 · Paper 2 · Question 9 of 20

TMUA 2017 Paper 2 Question 9

Proof and counterexample — Finding the first mistake in a proof. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 2Proof and counterexampleFinding the first mistake in a proof7 options
Consider the following attempt to prove this true theorem:

Theorem: a3 + b3 = c3 has no solutions with a, b and c positive integers.

Attempted proof:
Suppose that there are positive integers a, b and c such that a3 + b3 = c3.

I     We have a3 = c3 − b3.
II    Hence a3 = (c−b)(c2 + cb + b2).
III   It follows that a = c − b and a2 = c2 + cb + b2, since a ≤ a2 and c − b ≤ c2 + cb + b2.
IV   Eliminating a, we have (c−b)2 = c2 + cb + b2.
V    Multiplying out, we have c2 − 2cb + b2 = c2 + cb + b2.
VI   Hence 3cb = 0 so one of b and c is zero.

But this is a contradiction to the original assumption that all of a, b and c are positive. It follows that the equation has no solutions.

Comment on this proof by choosing one of the following options:

  1. AThe proof is correct
  2. BThe proof is incorrect and the first mistake occurs on line I.
  3. CThe proof is incorrect and the first mistake occurs on line II.
  4. DThe proof is incorrect and the first mistake occurs on line III.
  5. EThe proof is incorrect and the first mistake occurs on line IV.
  6. FThe proof is incorrect and the first mistake occurs on line V.
  7. GThe proof is incorrect and the first mistake occurs on line VI.
Show the answer and worked solution
answer · D
  1. AThe proof is correct
  2. BThe proof is incorrect and the first mistake occurs on line I.
  3. CThe proof is incorrect and the first mistake occurs on line II.
  4. DThe proof is incorrect and the first mistake occurs on line III.
  5. EThe proof is incorrect and the first mistake occurs on line IV.
  6. FThe proof is incorrect and the first mistake occurs on line V.
  7. GThe proof is incorrect and the first mistake occurs on line VI.
Lines I and II are just rearranging and using the difference of two cubes, which is fine. Line III is the leap: knowing that a3 equals the product of two positive integers does not tell you that those two factors are a and a2. The factorisation of a3 could split in many other ways — for instance a = 6 gives 216 = 4 × 54 — and the size comparison quoted proves nothing about which factor is which. Everything after line III is correct algebra applied to a false statement.