Equation : $5^{2(x+1)} +621\times{10}^x=100\times4^x$

Question

Find all integer roots of the equation : $$5^{2(x+1)} +621\times{10}^x=100\times4^x$$

It's easily seen that $x=0$ doesn't work.I feel there's no positive root for this equation but how to proceed?

Answer 1

Hint :

Divide both sides by $4^x$

$$25\left\{\left(\dfrac52\right)^x\right\}^2+621\left(\dfrac52\right)^x-100=0$$ which is a Quadratic Equation in $\left(\dfrac52\right)^x$

Now if $\displaystyle u^m=1,$

either $\displaystyle m=0,u\ne0; $

or $\displaystyle u=1$

or $\displaystyle u=-1,m$ is even

Answer 2

If $x\ge1$, the left hand side of $5^{2(x+1)}+621\times10^x=100\times4^x$ is odd while the right hand side is even, which is impossible, so we must have $x\le0$. Writing $x=-u$, and multiplying both sides by $2^{2u}\times5^{2u}$ to make all the powers non-negative, the equation becomes

$$5^2\times2^{2u}+621\times10^u=2^2\times5^{2+2u}$$

or

$$621\times10^u=5^2(2^2\times5^{2u}-2^{2u})$$

The right hand side is divisible by $5^2$ but by no higher power of $5$. So $u=2$ is the only possible integer solution. And, indeed, $5^2(2^2\times5^4-2^4)=10^2(5^4-2^2)=10^2(625-4)=621\times10^2$. Thus $x=-2$ is the sole integer solution.