TMUA 2022 Paper 2 Q7 — Worked Solution

P2·Number Theory & Divisibility, Logic & Quantifiers·6.5Official · 2022 · Q7

proof-errorirreducibility-vs-primalitycube-numbers

Consider the following claim:The difference between two consecutive positive cube numbers is always prime.Here is an attempted proof of this claim:

(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1

Taking

x

to be a positive integer, the difference between two consecutive cube numbers can be expressed as

(x + 1)^{3} - x^{3} = 3 x^{2} + 3 x + 1

III

It is impossible to factorise

3 x^{2} + 3 x + 1

into two linear factors with integer coefficients because its discriminant is negative.

Therefore for every positive integer value of

x

the integer

3 x^{2} + 3 x + 1

cannot be factorised.

Hence, the difference between two consecutive cube numbers will always be prime.

Which of the following best describes this proof?

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