If x and y are negative integers such that x = x^y + y, what is x?

\(x = x^y + y\) or \(x-x^y = y\)

Statement 1: this implies that \(y = -2\). putting the value of \(y\) in question stem equation we get

\(x-x^{-2}=-2\)

\(x-\frac{1}{x^2} = -2\), as \(x\) is a negative integer, only \(x=-1\) satisfies this equation. Hence sufficient

Statement 2: Directly provides the value of \(y\). As explained above only \(x=-1\) satisfies the equation. Hence sufficient

Option D

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Factorization to find the value of \(x\)

\(x-\frac{1}{x^2} = -2\) or \(x^3+2x^2-1=0\)

here \(f(x)=x^3+2x^2-1\), Notice that \(f(-1)=0\), or \(x=-1\) is a root of the equation, so \(x+1=0\) is a factor of the equation.

Divide the equation by \((x+1)\) to get \(x^3+2x^2-1=(x+1)(x^2+x-1) =0\),

Now \(x\) is a negative integer, so for any value of \(x\), \(x^2+x-1\) will not be \(0\)

Hence \(x+1=0\) or \(x=-1\)