Any way to show that $x=2$ is only integer solution of this equation :$x^x-x^2(1-x)^4=0$?

Question

$ x=2$ is clear that is a integer solution of this equation $x^x-x^2(1-x)^4=0$ , but to prove it it is hard by standard methods , I want if there is any method using variable change help me to solve this equation and to show that $x=2$ is only integer solution of this equation :$x^x-x^2(1-x)^4=0$? ?

Answer 1

Hint If $x >2$ then your equation is $$x^{x-2}=(1-x)^4$$ and the two sides are relatively prime integers.

If $x <2$ your equation becomes $$1=x^{2-x}(1-x)^4$$ and the RHS is the product of two integers.

Answer 2

$x^x-x²(1-x)^4=0$

So, $x^2 = \frac{x^x}{(1-x)^4}$

Since $x^2$ and $(1-x)^4$ are perfect squares, $x^x$ must also be a perfect square.

$x$ can't be 0

$x$ can only take values from 1 to 5 out of which $x^x$ is a perfect square only for 1, 2 and 4.

After substituting we can notice that only $2$ works.

Hence, $x$ = 2

Answer 3

For $x\ge6$, $$x^x\ge x^6>x^2(1-x)^4$$ and there cannot be a solution.

You conclude by checking with the values $0$ to $5$. (And obviously negative integers can't do.)

Answer 4

This equation is equivalent to $$x^x=x^2(1-x)^4. $$ If, $x\geq2$ then $$ x^{x-2}=(1-x)^4$$ and both sides are integers. We know $x=2$ is a solution to the equation, and if $x>2$, then $x$ divides the LHS of the equation, and therefore must divide the RHS, but $\gcd(x,1-x)=1$, which it can't happen. We know $x=1$ and $x=0$ is not a solution, and if $x<0$ the LHS of the first equation is not an integer, but the RHS is, hence they cannot be equal.