Consider the following claim:
The difference between two consecutive positive cube numbers is always prime.
Here is an attempted proof of this claim:
I (x+1)3 = x3 + 3x2 + 3x + 1
II Taking x to be a positive integer, the difference between two consecutive cube numbers can be expressed as (x+1)3 − x3 = 3x2 + 3x + 1
III It is impossible to factorise 3x2 + 3x + 1 into two linear factors with integer coefficients because its discriminant is negative.
IV Therefore for every positive integer value of x the integer 3x2 + 3x + 1 cannot be factorised.
V Hence, the difference between two consecutive cube numbers will always be prime.
Which of the following best describes this proof?
- AThe proof is completely correct, and the claim is true.
- BThe proof is completely correct, but there are counterexamples to the claim.
- CThe proof is wrong, and the first error occurs on line I.
- DThe proof is wrong, and the first error occurs on line II.
- EThe proof is wrong, and the first error occurs on line III.
- FThe proof is wrong, and the first error occurs on line IV.
- GThe proof is wrong, and the first error occurs on line V.
Show the answer and worked solution
answer · F
- AThe proof is completely correct, and the claim is true.
- BThe proof is completely correct, but there are counterexamples to the claim.
- CThe proof is wrong, and the first error occurs on line I.
- DThe proof is wrong, and the first error occurs on line II.
- EThe proof is wrong, and the first error occurs on line III.
- FThe proof is wrong, and the first error occurs on line IV.
- GThe proof is wrong, and the first error occurs on line V.
Lines I to III are sound — the discriminant 9 − 12 = −3 really is negative, so the quadratic has no linear factors over the integers. Line IV is the error: a polynomial being irreducible says nothing about whether its integer values are composite. Taking x = 7 gives 3(49) + 21 + 1 = 169 = 132, which is not prime, so the claim itself fails too.