Prove that if $p$ is prime and $a^7-b^3=p^2$ then $\text{gcd}(a,b)=1$

Question

Prove that if $p$ is prime and $a^7-b^3=p^2$ then $\text{gcd}(a,b)=1$.

Any help is appreciated.

Answer 1

If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.

So there is no prime that divides $a$ and $b$ and thus $\gcd (a,b)=1$.

Answer 2

Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....

What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.

For each of these choices what do you get when you look at $\frac {a^7}{hcf(a,b)} - \frac {b^3}{hcf(a,b)} = \frac {p^2}{hcf(a,b)}$?

Which of those choices will allow this to be true?

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Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.

So $a^6*\frac {a}{hcf(a,b)} - b^2\frac {b}{hcf(a,b)} = \frac {p^2}{hfc(a,b)}$. So that means $\frac {p^2}{hfc(a,b)}$ is also a multiple of $hcf(a,b)$.

So $a^5 (\frac {a}{hcf(a,b)} )^2 - b(\frac {b}{hcf(a,b)})^2 = \frac {p^2}{hcf^2(a,b)}$. So that means that $\frac {p^2}{hcf^2(a,b)}$ is also a multiple of $hcf(a,b)$.

So $a^5(\frac {a}{hcf(a,b)} )^3- (\frac {b}{hcf(a,b)})^3 = \frac {p^2}{hcf^3(a,b)}$ is an integer.

How can $\frac {p^2}{hcf^3(a,b)}$ possibly be an integer?