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Prove that $\forall x \in \mathbb{N}, x^2 + 5x + 4$ is composite.

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    Note that your answer needs to exclude the possibility that one of the factors may be $1$ - or to deal with this case separately (e.g. $1\times 4=4$ is composite because $4=2\times 2$) – 2017-02-15

3 Answers 3

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Factorise $x^2 + 5x + 4$ into $(x+1)(x+4)$. For $x$ even note that one factor is odd and other even. For $x$ odd one factor is even and other odd. Thus always is even and composite.

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The polynomial can be rewritten as $$(x+1)(x+4)$$ which is in $\mathbb{N}$ for any $x \in \mathbb{N}$. Now observe, that both factors are larger than $1$ for any natural number $x \in \mathbb{N}$. Hence $$x^2 + 5x + 4$$ has other divisors than $1$ and itself, hence it is composite. This is taken from the second line here.

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Hint: $$x^2 + 5x + 4 = (x+1)(x+4).$$

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    I know but how do we proceed from there? Essentially proving it to be true using the definition of a composite number. – 2017-02-15
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    $x^2 + 5x + 4 = (x+1)(x+4).$ is composite since it has factors $(x+4)$ and $(x+1)$, and $x+4>4$ with a natural numbered value. – 2017-02-15
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    @HKT: What IS your definition of "composite" if $(x+1)(x+4)$ does not _immediately_ fit it? – 2017-02-15