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've already found $d$ divides $5m+3$
Again, $d$ divides $4m(5m+3)-5(4m^2+1)=12m-5$
$\implies d$ divides $12(5m+3)-5(12m-5)=61$
Now, $4m^2+1\equiv0\pmod{61}\iff m\equiv\pm36\pmod{61}$
