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The number of integers that satisfy the equality $$(x^{2}-5x+7)^{x+1}=1$$ is
$$\left(x^2-5x+7\right)^{x+1}=1$$
There can be a solution when $$\left(x^2-5x+7\right)=1$$ or $$x^2-5x\ +6=0$$
or x=3 and x=2
There can also be a solution when x+1 = 0 or x=-1
Hence three possible solutions exist.
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