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TMUA 2019 · Paper 2 · Question 7 of 20

TMUA 2019 Paper 2 Question 7

Proof and counterexample — Choosing the statement a given example refutes. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 2Proof and counterexampleChoosing the statement a given example refutes4 options
For which one of the following statements can the fact that 122 + 162 = 202 be used to produce a counterexample?
  1. AIf a, b and c are positive integers which satisfy the equation a2 + b2 = c2, and the three numbers have no common divisor, then two of them are odd and the other is even.
  2. BThe equation a4 + b2 = c2 has no solutions for which a, b and c are positive integers.
  3. CThe equation a4 + b4 = c4 has no solutions for which a, b and c are positive integers.
  4. DIf a, b and c are positive integers which satisfy the equation a2 + b2 = c2, then one is the arithmetic mean of the other two.
Show the answer and worked solution
answer · B
  1. AIf a, b and c are positive integers which satisfy the equation a2 + b2 = c2, and the three numbers have no common divisor, then two of them are odd and the other is even.
  2. BThe equation a4 + b2 = c2 has no solutions for which a, b and c are positive integers.
  3. CThe equation a4 + b4 = c4 has no solutions for which a, b and c are positive integers.
  4. DIf a, b and c are positive integers which satisfy the equation a2 + b2 = c2, then one is the arithmetic mean of the other two.
Work out what the numbers 12, 16, 20 actually do. They share the common divisor 4, so the first statement's hypothesis is not met and it cannot be refuted by them. For the last statement, 16 = 12 + 202, so the conclusion holds and again there is no counterexample. The one that works is the second: 162 = 28 = 44, so 44 + 122 = 256 + 144 = 400 = 202, giving positive integers a = 4, b = 12, c = 20 with a4 + b2 = c2. The third fails because 122 and 202 are not fourth powers.