TMUA 2017 Paper 2 Q9 — Worked Solution

P2·Algebra & Surds, Number Theory & Divisibility·7.0Official · 2017 · Q9

proof-errorfactorisation-not-uniquefermat-like

Consider the following attempt to prove this true theorem:Theorem:

a^{3} + b^{3} = c^{3}

has no solutions with

a

b

and

c

positive integers.Attempted proof:
Suppose that there are positive integers

a

b

and

c

such that

a^{3} + b^{3} = c^{3}

We have

a^{3} = c^{3} - b^{3}

Hence

a^{3} = (c - b) (c^{2} + c b + b^{2})

III

It follows that

a = c - b

and

a^{2} = c^{2} + c b + b^{2}

, since

a \leq a^{2}

and

c - b \leq c^{2} + c b + b^{2}

Eliminating

a

, we have

(c - b)^{2} = c^{2} + c b + b^{2}

Multiplying out, we have

c^{2} - 2 c b + b^{2} = c^{2} + c b + b^{2}

Hence

3 c b = 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.

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