If (x^2+5x+5)^{x+5}=1,then the number of inte | vedclass

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(C) The equation $(x^2+5x+5)^{x+5}=1$ holds true in the following cases:Case $1$: The exponent is $0$ and the base is non-zero.$x+5=0 \Rightarrow x=-5

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$4$

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(C) The equation $(x^2+5x+5)^{x+5}=1$ holds true in the following cases:Case $1$: The exponent is $0$ and the base is non-zero.$x+5=0 \Rightarrow x=-5$.Checking the base: $(-5)^2+5(-5)+5 = 25-25+5 = 5 
eq 0$. Thus,$x=-5$ is a solution.Case $2$: The base is $1$.$x^2+5x+5=1$ $\Rightarrow x^2+5x+4=0$ $\Rightarrow (x+1)(x+4)=0$.So,$x=-1$ and $x=-4$ are solutions.Case $3$: The base is $-1$ and the exponent is an even integer.$x^2+5x+5=-1$ $\Rightarrow x^2+5x+6=0$ $\Rightarrow (x+2)(x+3)=0$.So,$x=-2$ or $x=-3$.For $x=-2$,the exponent is $x+5 = -2+5 = 3$ (odd),so this is not a solution.For $x=-3$,the exponent is $x+5 = -3+5 = 2$ (even),so $x=-3$ is a solution.The set of integer solutions is $\{-5, -1, -4, -3\}$.Therefore,the number of integers satisfying the equation is $4$.

If $(x^2+5x+5)^{x+5}=1$,then the number of integers satisfying this equation is

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