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 $a ^ { 3 } + b ^ { 3 } = c ^ { 3 }$.
I We have $a ^ { 3 } = c ^ { 3 } - b ^ { 3 }$.
II $\quad$ Hence $a ^ { 3 } = ( c - b ) \left( c ^ { 2 } + c b + b ^ { 2 } \right)$.
III It follows that $a = c - b$ and $a ^ { 2 } = c ^ { 2 } + c b + b ^ { 2 }$, since $a \leqslant a ^ { 2 }$ and $c - b \leqslant c ^ { 2 } + c b + b ^ { 2 }$.
IV Eliminating $a$, we have $( c - b ) ^ { 2 } = c ^ { 2 } + c b + b ^ { 2 }$.
V Multiplying out, we have $c ^ { 2 } - 2 c b + b ^ { 2 } = c ^ { 2 } + c b + b ^ { 2 }$.
VI 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:
A The proof is correct
B The proof is incorrect and the first mistake occurs on line I.
C The proof is incorrect and the first mistake occurs on line II.
D The proof is incorrect and the first mistake occurs on line III.
E The proof is incorrect and the first mistake occurs on line IV.
F The proof is incorrect and the first mistake occurs on line V.
G The proof is incorrect and the first mistake occurs on line VI.