examday

/TMUA

A student attempts to prove the statement

For all primes $p>3$, $p^3-3p^2$ has at least 12 factors.

Their proof is as follows.

• (I) $p^3-3p^2=p^2(p-3)$.
• (II) As $p>3$, $p$ is odd, therefore $p-3=2a$ for some integer $a\ne0$, therefore $p^2(p-3)=p^2\cdot2\cdot a$.
• (III) If $a=p$, then $p=-3$ which is impossible, so $a\ne p$.
• (IV) If $N$ has prime factorisation $p_1^{a_1}\cdot p_2^{a_2}\cdots p_n^{a_n}$, then each factor of $N$ has the form $p_1^{b_1}\cdot p_2^{b_2}\cdots p_n^{b_n}$, where $0\le b_i\le a_i$ (for $1\le i\le n$).
• (V) If $a$ is prime, then $p^2\cdot2^1\cdot a^1$ has $(2+1)(1+1)(1+1)=12$ factors.
• (VI) If $a$ is composite, then $p^2\cdot2^1\cdot a=p^2\cdot2^1\cdot q^1\cdot b$ for some prime $q$ and integer $b>1$, which has more than 12 factors.

Which of the following best describes the proof?