Are all those numbers coprime? The values of $4m^2+1$ and $4m^2+4m+5$ for $m\geq{1}$ are (resp.) 5,17,37,... and 13,29,53,... Those numbers seem to be all coprime : how to prove it if it is true, please ? 
 A: By inspection, the two numbers are not relatively prime if $m=-1$, for both are equal to $5$. 
It follows that the two numbers are both divisible by $5$ if $m\equiv -1\pmod{5}$.  
A: Suppose there is a common divisor $d$. Then $d|4m^2+1$ and $d|4m^2+4m+5$ implies that $d|4m^2+4m+5-4m^2-1$, which is $d|4m+4$. Since $4m^2+1$ is an odd number, so is $d$. Therefore $d|m+1$. Also, $d|4(m+1)^2$, so $d|4(m+1)^2-(4m^2+4m+5)$ and then $d|4m-1$.
Now combine $d|4m-1$ and $d|4m+4$. This gives us $d|5$.
A: Suppose that $\,d\mid \smash[b]{\underbrace{4m^2\!+1}},\ $ and $\,d\mid f(m)  = \color{#0a0}{4m^2}\!+\!4m\!+\!5.\ $ We prove that $\ \color{#c00}{d\mid 5}.$
Note $\ {\rm  mod}\ d:\,\ \color{#0a0}{4m^2}\equiv -1,\,$ hence $\  f(m) \equiv 4(m\!+\!1)\equiv 0$.
Hence, scaling by $\,1/4\equiv -m^2$ yields that $\  m+1\equiv 0\,\Rightarrow\,m\equiv -1\,\Rightarrow\, 0\equiv f(m)\equiv f(-1)\equiv \color{#c00}5$
A: So you're asking if $\gcd(4m^2 + 1, 4m^2 + 4m + 5) = 1$ always. If you had just looked at $m = 4$, you would noticed $65$ and $85$, and maybe you would also have been led to the observation that $\gcd(4m^2 + 1, 4m^2 + 4m + 5) = 5$ can also occur.
Compute $4m^2 + 1 \pmod 5$ to get the periodic sequence $0, 2, 2, 0, 1, \ldots$
Now compute $4m^2 + 4m + 5 \pmod 5$ to get the periodic sequence $3, 4, 3, 0, 0, \ldots$
So if $m \equiv 4 \pmod 5$, then $4m^2 + 1 \equiv 0 \pmod 5$ and $4m^2 + 4m + 5 \equiv 0 \pmod 5$ as well.
