Let's start by assuming that \(\frac{(x^2 + 1)}{(x + 5)}\) is an even number. Therefore \(\frac{(x^2 + 1)}{(x + 5)}=2p\) for some integer \(p\). Cross-multiplying and forming a quadratic, we get:
\(x^2-2px+1-10p=0\)
Solving for \(x\) using the Quadratic formula, we get \(x = p ± \sqrt{p^2+10p-1}\)
Now since we have been given that \(x\) is an integer, it is necessary that \(\sqrt{p^2+10p-1}\) is an integer as well. Let's assume \(p^2+10p-1=m^2\) for some integer \(m\). Rearranging and forming a quadratic, we get:
\(p^2+10p-1-m^2=0\)
Solving for \(p\) using the Quadratic formula, we get \(p = -5 ± \sqrt{26+m^2}\)
But since we had assumed initially that \(p\) is an integer, therefore \(\sqrt{26+m^2}\) must also be an integer. Let's assume \(26+m^2=k^2\) for some integer \(k\). Rearranging, we get:
\(k^2-m^2=26\) which can further be written as \((k+m)(k-m)=26\) where both \(k\) and \(m\) are integers.
In order for this to hold, \((k+m)\) and \((k-m)\) must be of the same even/odd nature. But from factorization of \(26\) we know that \(26\) cannot ever have a pair of integral factors such that they are of the same even/odd nature. (\(26\) and \(1\) are different in their even/odd nature, and so are \(2\) and \(13\))
Thus,
1. There is no \(k\) for which \(\sqrt{26+m^2}\) is an integer. This implies that
2. There is no \(m\) for which \(\sqrt{p^2+10p-1}\) is an integer. This implies that
3. There is no \(p\) for which \(\frac{(x^2 + 1)}{(x + 5)}\) is an integer.
Thus, this contradicts our initial assumption and proves that for no \(x\) can \(\frac{(x^2 + 1)}{(x + 5)}\) ever be an even integer.
We can confidently say that Statement 1 is sufficient.
From here it is a simple step to observe that all elements in Statement 2 are indeed a subset of elements in Statement 1 and thus our conclusion holds true for Statement 2 as well.
Therefore, either statement is sufficient in answering the question (the answer is NO, the given fraction cannot be an even integer for any integer \(x\)) and hence option D is our correct option choice for this question.This brings us to the end of this super long post. I hope you guys had fun!

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