Question

The number of integers that satisfy the equality $(x^2-5x+7)^{x+1} = 1$ is

• A. 2
• B. 3
• C. 5
• D. 4

Answer 1

The Correct Option is B

The correct answer is (B): $3$

$(x^2-5x+7)^{x+1} = 1$

We know, for $a^b=1$, if

$-a =-1$ then $b$ is even.

$-a = 1$ then $b$ is any number 

$-a>0$ then $b=0$

Case 1: $x+1=0 ⇒ x = -1$

Case 2: $x^2-5x+7 = 1 ⇒ x^2-5x+6=0 ⇒ x = 2 \;or \;3$

Case 3: $x^2-5x+7 = -1 ⇒ x^2-5x+8=0$

but $x$ is not an integer 

$∴$ The number of integers satisfies the equation is $3$