Page 1 of 1

no solution (a,b)

Posted: Sat Jul 01, 2017 3:40 pm
Prove that there is no pair $(a,b)$ of integers such that
$a^2=b^7+7$

Re: no solution (a,b)

Posted: Tue Oct 16, 2018 8:01 pm
The LHS of the equation is always positive. So $b>0$, as for $b=0, a= \sqrt7$, and when $b=-1, a=\sqrt6$. And when $b<-1$, the equation becomes invalid. Therefore, it can be deduced that $b>0$.

$a^2-b^7=7$
$(a+\sqrt{b^7}) (a-\sqrt{b^7})=7$

Now, $a>b$, otherwise the equation draws into a negative result on the LHS. And $a$ is considerably greater than $b$, like when $b=2, \sqrt{b^7}=\sqrt128= 8\sqrt2$, so $a>12$, as otherwise, the equation would become negative. But there is no way that a value of $7$ could be obtained under this circumstances, and as the value of $b$ increases, the value on the LHS drifts further away from $7$.

So, by deduction, it can be concluded that the equation $a^2=b^7+7$ has no pair $(a,b)$ of integers solution.

Re: no solution (a,b)

Posted: Fri Oct 19, 2018 1:43 am
There can be another solution to this problem, which includes a bit of messy work of Algebra.

$a^2=b^7+7$
$a^2-16=b^7-9$
$(a+4)(a-4)=(\sqrt{b^7}+3)(\sqrt{b^7}-3)$

Now using a little bit of Algebra, it can be shown that there is no integer value of $a$ or, $b$ that satisfies the above equation, hence the proof.
Anyone can do it by themselves, so I'm saving myself from the hazard of typing the long and messy algebraic proof.