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+## Part 2
+
+>
+> Show that for all integers $n$, if $n$ is a multiple of 27, then $n+5$ and $n^2+2$ are coprime.
+
+2 expressions (for example $a$ and $b$) are co-prime if $gcd(a,b)=1$.
+
+By the division algorithm, if $a=bq+r$, then $gcd(a,b)=gcd(b,r)$.
+
+Consider that $n^2+2$ can be expressed in the form $n^2-25+27$, which in turn can be factorised as $(n+5)(n-5)+27$ through application of the "difference of two squares" identity.
+
+If $n$ is a multiple of 27, $n=27k$, where $k\in\mathbb{Z}$. As a consequence, $n+5=27k+5\space(1)$.
+
+Using the division algorithm, $gcd(n^2+2,n+5)=gcd(n+5,27)\space(2)$.
+
+Given $(1)$, equation $(2)$ can be simplified to $gcd(5,27)$. 5 is a prime number, and not a factor of 27, meaning the GCD must be 1.
+
+Since the greatest common divisor of $n^2+2$ and $n+5$ is 1, the two expressions are co-prime.
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