#1**+1 **

We can use Euclid's algorithm for a few steps:

gcd(2π2+29π+65,π+13)=gcd(2π2+29π+65β2π(π+13),π+13)=gcd(3π+65,π+13)=gcd(3π+65β3(π+13),π+13)=gcd(26,π+13)gcd(2a2+29a+65,a+13)=gcd(2a2+29a+65β2a(a+13),a+13)=gcd(3a+65,a+13)=gcd(3a+65β3(a+13),a+13)=gcd(26,a+13)

which is necessarily 1,2,131,2,13 or 2626, just by looking at the first term. By factoring 11831183, and considering that πa is an odd multiple, you should be able to conclude the rest

I'm sorry for the bad spacing I wrote this in apple notes and copy-pasted

Here to help

:D

LuckyDucky Apr 29, 2020

#1**+1 **

Best Answer

We can use Euclid's algorithm for a few steps:

gcd(2π2+29π+65,π+13)=gcd(2π2+29π+65β2π(π+13),π+13)=gcd(3π+65,π+13)=gcd(3π+65β3(π+13),π+13)=gcd(26,π+13)gcd(2a2+29a+65,a+13)=gcd(2a2+29a+65β2a(a+13),a+13)=gcd(3a+65,a+13)=gcd(3a+65β3(a+13),a+13)=gcd(26,a+13)

which is necessarily 1,2,131,2,13 or 2626, just by looking at the first term. By factoring 11831183, and considering that πa is an odd multiple, you should be able to conclude the rest

I'm sorry for the bad spacing I wrote this in apple notes and copy-pasted

Here to help

:D

LuckyDucky Apr 29, 2020