Prove that there is no pair $(a,b)$ of integers such that
$a^2=b^7+7$
no solution (a,b)

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: no solution (a,b)
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 $b>0$.
$a^2b^7=7$
$(a+\sqrt{b^7}) (a\sqrt{b^7})=7$
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.
$a^2b^7=7$
$(a+\sqrt{b^7}) (a\sqrt{b^7})=7$
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.

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: no solution (a,b)
There can be another solution to this problem, which includes a bit of messy work of Algebra.
$a^2=b^7+7$
$a^216=b^79$
$(a+4)(a4)=(\sqrt{b^7}+3)(\sqrt{b^7}3)$
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.
$a^2=b^7+7$
$a^216=b^79$
$(a+4)(a4)=(\sqrt{b^7}+3)(\sqrt{b^7}3)$
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.