Modular calculus and square I want to prove that $4m^2+1$ and $4m^2+5m+4$ are coprimes and also $4m^2+1$ and $4k^2+1$ when $k\neq{m}$ and $4m^2+5m+4$ and $4k^2+5k+4$ when $k\neq{m}$. Firstly : Let $d|4m^2+1$ and $d|4m^2+5m+4$ then $d|4m^2+5m+4-(4m^2+1)=5m+3$ and $d|5m^2+3m$ thus $d|5m^2+3m-(4m^2+5m+4)=m^2-2m-4$ and $d|4m^2-8m-16$ or $d|4m^2+5m+4-(4m^2-8m-16)=13m+20$ but $d|15m+9$ thus $d|15m+9-(13m+20)=2m-11$ but $d|13m+20-(5m+3)=8m+17$ and $d|8m-88$ therefore $d|8m+17-(8m-88)=105=5.3.7$. So $4m^2+1=105k$ or $4m^2=104+105k'$ but $k'=4k''$ then $m^2=26+105k''$ or $4m^2+1=21k$ or $4m^2+1=35k$ or $4m^2+1=15k$ or $4m^2+1=5k$ or 3k or 7k. Are there solutions to these equations, please, or can I conclude that the two numbers are coprimes ?

You have an error in your calculation.
$8m-88$ should be $4(2m-11)=8m-44$. Then, $(8m+17)-(8m-44)=61$.
The two numbers are not always coprime. For example, for $m=36$, 
$$4m^2+1=5185=85\cdot 61,\ \ 4m^2+5m+4=5368=88\cdot 61.$$
In general, for $m=61k-25$ where $k$ is a positive integer, the two numbers have $61$ as a common prime factor :
$$4m^2+1=61(244k^2-200k+41),\ \ \ 4m^2+5m+4=61(244k^2-195k+39).$$
Also, for $m\gt 0,k\gt 0,m\not=k$, since
$$4m^2+1=4k^2=1\iff 4(m-k)(m+k)=0,$$
we have $4m^2+1\not =4k^2+1$.
For $m\gt 0,k\gt 0,m\not=k$, since
$$4m^2+5m+4=4k^2+5k+4\iff (m-k)(4m+4k+5)=0,$$
we have $4m^2+5m+4\not=4k^2+5k+4$.
