Exercise 1.3.33 ($n^4 + n^2 + 1$ is composite)

Question

Show that $n^4+n^2 + 1$ is composite if $n>1$.

Richard Ganaye · Accepted Answer

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

Exercise 1.3.33 ($n^4 + n^2 + 1$ is composite)

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Exercise 1.3.33 ($n^4 + n^2 + 1$ is composite)

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