Exercise 1.3.32 ($n^4 + 4$ is composite.)

Question

Show that $n^4+4$ is composite for all $n>1$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $N = n^4 + 4$, with $n>1$. Then
\begin{align*}
N &= n^4+4\
&= (n^2 + 2)^2 - 4n^2\
&= (n^2 - 2n +2)(n^2 + 2n+2).
\end{align*}
Moreover, $n^2 - 2n + 2 = (n-1)^2 + 1 >1$ if $n>1$. A fortiori $n^2 + 2n+2 >1$. This proves that 
$$n^4 + 4 = (n^2 - 2n +2)(n^2 + 2n+2)$$
is composite for $n>1$.

\end{proof}

Exercise 1.3.32 ($n^4 + 4$ is composite.)

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Exercise 1.3.32 ($n^4 + 4$ is composite.)

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