Question

Let m be a positive integer.Find all pairs of integers (x,y) such that $x^2(x^2+7)=y^{m+1}.$

Answer 1

Given equation can be written as

$x^2(x^2+7)=y^{m+1}\times 1$

There are two possibilities for the equation to be true, written as follows

$i)x^2=1\Rightarrow x=\pm 1$

Now,

$x^2+7=y^{m+1}$ [ When $x=1\Rightarrow x^2=1$]

$y^{m+1}=1+7\Rightarrow 2^3$

Since $m$ is a positive integer therefore substituting $m=2$ in above equation, we get

$y^{2+1}=2^3$

On comparing, we get

$y=2.$

When $x=-1$

Now,

$x^2+7=y^{m+1}$

$y^{m+1}=1+7\Rightarrow 2^3$ [When $x=-1\Rightarrow x^2=1$]

Since $m$ is a positive integer therefore substituting $m=2$ in above equation, we get

$y^{2+1}=2^3$

On comparing, we get

$y=2.$

$ii)x^2+7=1\Rightarrow x^2=-6$

Since being complex values of $x$ therefore the solution does not exist for this possibility.

Hence, the pairs $(x,y)=(1,2),(-1,2)$ satisfy the given equation at $m=2$ when $m$ is an arbitrary positive integer.