Are you sure you have all values?

$\text{Case 1:}$

$x^2 - 11x + 30=0$

$(x-5)(x-6)=0\Rightarrow x=5,6$

$\text{Case 2:}$

$x^2 - 5x + 5=1$

$x^2 - 5x + 4=0$

$(x-1)(x-4)=0\Rightarrow x=1,4$

$\text{Case 3:}$

$x^2 - 5x + 5=-1$

$x^2 - 5x + 6=0$

$(x-3)(x-2)=0\Rightarrow x=2,3$

$\color{#333333}{\text{Important}}$

Then check that $x^2 - 11x + 30$ is even when substitute $x$ by $2$ and $3$ . By this cases the values of $x^2 - 11x + 30$ are $72$ and $56$ , respectively.

$\therefore \boxed{x=1,2,3,4,5,6}$

The LHS will be equal to 1 when either the base=1 ie. $x^{2} - 5x + 5 = 1$ so $x^{2}-5x+4=0 >>>$ which gives x=4 and x=1 , Or when the power is 0 ie. $x^{2}-11x+30=0 >>> x^{2}-5x-6x+30>>>$ which gives x=5 and x=6 , Or when the base is(-1) and the power is even so $x^{2}-5x+5=-1 >>> x^{2}-5x+6=0$ which gives x=2 and x=3 but x=2 and x=3 on substituting in the power should be even which is true hence x={1,2,3,4,5,6} hence 6 is the answer, also in the second case notice that the base has will be equal to 0 only for irrational values so the situation $0^{0}$ can never occur for integer values

Their are basically three cases in which the result will be 1 ie. $a^n=1$ if Case 1 a=any integer and n is zero Case 2 a=1 and be any integer Case 3 $a=-1$ and n is even integer Now let us proceed towards our problem Case 1: when $x^2-11x+30=0$ Or $(x-6)(x-5)=0$ Ie. $x=6,5$ Case 2: when $x^2-5x+5=1$ Or $(x-4)(x-1)=0$ ie. $x=4,1$ Case :3 when $x^2-5x+5=-1$ Or $(x-2)(x-3)=0$ ie. $x=3,2$ $**note**$ this case is only valid when value of the power is even,ie. When $x^2-11x+30$ is even. Let us see whether $x^2-11x+30=0$ is even or not,here when x is even then $(even)^2-(even)+(even)$ is always a even now in case of odd $(odd)^2-(odd)+(even)$ will again always even, so all the above 3 cases are are verified hence their are 6 values for which the above expression helds true which are $x=1,2,3,4,5,6$ so the answer is $\boxed{6}$

3 cases are possible for the question :

Case 1 :

We know that $1^x = 1$ for any real $x$ .

Hence, setting $x^2-5x+5=1$ ,

$x^{2}-5x+4=0$

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

$x=1 ,4$

Case 2:

We know that $x^0=1$ for any $x$ except $0$

Setting $x^2-11x+30=0$ , we get

$(x-5)(x-6) = 0$ .

$x=5,6$

Case 3:

$(-1)^n = 1$ for even $n$ .

Therefore, in addition to putting $x^2-5x+5 = -1$ , we also need to check if $x^2-11x+30$ is even.

$x^2-5x+6=0$

$(x-2)(x-3) = 0$

$x=2,3$

Substitution 2 and 3 in $x^2-11x+30$ , we notice that in both cases, it comes out to be even.

Hence, $6$ integers satisfy the given equation.

There are 3 different ways you can achieve 1. x^2-5x+5=1 2. x^2-11x+30=0 3. x^2-5x+5=-1 and x^2-11x+30= multiple of 2. Quadratic cases 1 and 2, you get 6, 5, 4, 1. For the last, you get 3 and 2 for x^2-5x+5=-1, and to confirm these you need to subtract from 121 to make the square of 5 and the square of 7. Subtract squares of 7 and 5 from 121, and you find that the result is divisible by 4.

1, 2, 3, 4, 5, 6 all work.

We'll denote the above expression as a^b=1. There are 3 cases to make the expression true. either: a = 1 (1 to any power will be 1) b = 0 (numbers to power of 1 will be 1) or a = -1 (as long as b is even) hence: x^2 -5x +5 = 1 x=1 or x=4 [these are two integers here] then: x^2-11x+30=0 x=6 or x=5 [these are two integers here] and: x^2-5x+5=-1 x=2 or x=3 [two integers*] since -1 to an even power is 1, we plugin our integer values into "b" and check that both integers make the case true.

As a total there are 6 different integers that satisfy the equation.