Prove gcd equality

Question

How to prove in Euclidean ring $Z$ with $a\in Z$ $$gcd(a^2+3, 3a+5)=gcd(a^2+3, 3a^2+5a, 3a+5)$$ ?

Answer 1

We have the following three results:

Let $a$ and $b$ be any integers not both zero. If $b \neq 0$ and $b$ divides $a$, then $$ \gcd (a, b) = b. $$

Let $a$ and $b$ be any integers such that $b \neq 0$ and $a = bq + r$ for some integers $q$ and $r$. Then $$ \gcd(a, b) = \gcd( b, r). $$

Let $a$, $b$, and $c$ be any integers at most one of which can be zero. Then we have $$ \gcd(a, b, c) = \gcd \big( a, \gcd(b, c) \big). $$

In our proof we will be using the above two results. So here is the proof:

We note that $$ \begin{align} \gcd\left( a^2 + 3, 3a^2 + 5a, 3a + 5 \right) &= \gcd \left( a^2 + 3, a(3a+5), 3a+5 \right) \\ &= \gcd \left( a^2 + 3, \gcd \big( a(3a+5), 3a+5 \big) \right) \\ &= \gcd \left( a^2 + 3, 3a+5 \right), \end{align} $$ as required.

Hope this helps.

Please feel free to ask for explanation / justification of whatever point you are unsure of.