# Solve for integer values of x. - - - (A^B = 1 form of equation)

• Aug 7th 2018, 02:24 PM
greg1313
Solve for integer values of x. - - - (A^B = 1 form of equation)
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Here is a problem I made up:

Let x belong to the set of integers.

That is, let x belong to the set containing the negative integers, zero, and the positive integers.

$\displaystyle Let \ \ A \ = \ (x \ + \ (x - 1)\sqrt{x^2 - 1 \ } \ ), \ \ and \ \ let \ \ B \ = \ (-3x^2 - x + 4)$.

Imaginary numbers will be allowed for the $\displaystyle \ \sqrt{x^2 - 1} \$ expression.

Solve $\displaystyle \ A^B = 1, \ \$ for all integer values of x only.
• Aug 7th 2018, 02:47 PM
SlipEternal
Re: Solve for integer values of x. - - - (A^B = 1 form of equation)
Find the values where $A=1$, where $B=0, A\neq 0$, or where $A=\phi, n|B$ where $\phi$ is an $n$-th root of unity:

$$1=x+(x-1)\sqrt{x^2-1}$$

If $x\neq 1$:

$$\sqrt{x^2-1}=-1$$

Only solution over integers is $x=1$.

$$-3x^2-x+4=0$$

$$(3x+4)(-x+1)=0$$

Only integer is $x=1$ again.

Finally, we try $A=\phi$:

$$\phi=x+(x-1)\sqrt{x^2-1}$$

Again, if $x\neq 1$ we have:

$$\left( \dfrac{\phi-x}{x-1} \right)^2=x^2-1$$

$$(x-\phi)^2=(x^2-1)(x-1)^2$$

Since the RHS is an integer, so too must the left hand side be an integer. This means that $x=-1$ or $x=0$.

Therefore the solution is $x=-1,0,1$.