Relevant wiki: Greatest Common Divisor - Problem Solving

Suppose that $m,n \in \mathbb{N}$ are such that $m > 1$ . If $d$ divides both $n+m$ and $mn+1$ , then $d$ divides $m(n+m)-(mn+1) = m^2-1$ , and hence $d$ divides both $n+(m^2-1)+m$ and $m(n+m^2-1)+1$ . In other words, $d$ divides $f_m(n+m^2-1)$ . Thus we deduce that $f_m(n)$ divides $f_m(n+m^2-1)$ . On the other hand, if $d$ divides both $n+(m^2-1)+m$ and $m(n+m^2-1)+1$ , then it also divides $m\big(n+(m^2-1)+m\big) - \big(m(n+m^2-1)+1\big) = m^2-1$ , and hence it divides both $n+m$ and $mn+1$ as well, so that it divides $f_m(n)$ . Thus we deduce that $f_m(n+m^2-1)$ divides $f_m(n)$ . In other words, we have shown that $f_m(n + m^2-1) = f_m(n)$ . This means that $T_m$ divides $m^2-1$ for all integers $m > 1$ .

Note that $f_m(m^2-m-1)$ is the highest common factor of $m^2-1$ and $m(m^2-m-1) + 1 = m^3-m^3-m+1 = (m-1)(m^2-1)$ , and so $f_m(m^2-m-1) = m^2-1$ . This means that $f_m(m^2-m-1+T_m) = m^2-1$ . This implies that $m^2-1$ divides $m^2-1+T_m$ , and hence that $m^2-1$ divides $T_m$ . Thus we have shown that $T_m = m^2-1$ for all integers $m > 1$ . $T_{12} = \boxed{143}$ .

---

An earlier version of the question asked for $\sum_{m=2}^\infty \frac{1}{T_m} \; = \; \sum_{m=2}^\infty \frac{1}{(m-1)(m+1)} \; = \; \frac12\sum_{m=2}^\infty \left(\frac{1}{m-1} - \frac{1}{m+1}\right) \; = \; \frac34$ making the answer $3+4=\boxed{7}$ .

Relevant wiki: Greatest Common Divisor - Problem Solving

Key fact : $\gcd(a,b) = \gcd(a,b + ka)$ for any integer $k$ . This will be used several times in the reasoning below.

First, note that $\gcd(m+n,mn+1) = \gcd\left(m+n\ ,\ m(m+n) - (m^2-1)\right) = \gcd(m+n,m^2-1)$ and likewise $\gcd(m+n+T,m(n+T)+1) = \gcd\left(m+n+T\ ,\ m(m+n+T) - (m^2-1)\right) = \gcd(m+n+T,m^2-1).$ We want to find a positive value for $T$ for which these two expressions are equal independent of the value of $n$ . Obviously, $T = m^2-1$ works because $\gcd(m+n+(m^2-1),m^2-1) = \gcd(m+n,m^2-1).$ But $T = m^2-1$ is also the smallest positive value that works. To see that, suppose $T < m^2-1$ . Choose $n = (m^2-1)-m$ . Then $\gcd(m+n,m^2-1) = \gcd(m^2-1,m^2-1) = m^2-1\ \ \ \ \text{but}\ \ \ \ \ \gcd(m+n+T,m^2-1) = \gcd(m^2-1+T,m^2-1) = \gcd(T,m^2-1) \leq T < m^2-1,$ which is a contradiction.

We conclude $\boxed{T_m = m^2-1}$ and $\boxed{T_{12} = 12^2-1 = 143}$ .

Relevant wiki: Greatest Common Divisor - Problem Solving

$gcd ( n+m, mn+1 ) = gcd (n+m, mn+1 - m(n+m) ) = gcd ( n+m, 1-m^2) = gcd ( n+m, m^2-1)$ . Clearly this is periodic in $n$ with a period of $m^2-1$ , but we must show that this is the fundamental period. $gcd(n+m, m^2-1) = m^2-1$ when $m^2-1 | n+m$ , meaning that this value cannot occur more frequently than every $m^2-1$ values of $n$ , proving that the period is fundamental. Lastly, $12^2-1 = 143$

This is a rusty C program,but it works.Here in main()," i" acts as period T and assuming T<100000 we get the answer 143

include<stdio.h>

int gcd(int a,int b)
{
    if(a>b)
    {
        a=a-b;
        gcd(a,b);
    }
    else if(a<b)
    {
        b=b-a;
        gcd(a,b);
    }
    else if(a==b)
    {
        return a;
    }
}
int fn(int i)
{
    int f=gcd(12+i,12*i+1);
    return f;
}
void main()
{
    int i,j,flag=0;
    for(i=1;i<100000;i++)
    {
        for(j=1;j<100;j++)
        {
             if( fn(i+j)==fn(j))
             {
                 flag=1;
             }
             else
             {
                 flag=0;
                 break;
             }
        }
    if(flag==1)
        {
            printf("%d",i);
            break;
        }
        flag=0;
    }
}

$f_m(n) = gcd(m+n,mn+1)$ $\Rightarrow f_m(n) = gcd((m-1)(n-1),mn+1 -------------(1)$ for a period $T_m$ we have , $\Rightarrow f_m(n+T_m) = gcd((m-1)(n+T_m-1),mn+mT_m+1) = d (say) ----------------------------(2)$ so, $d|(m+n)-------(*)$ and $d|mn+1--------(*)$ implies $\Rightarrow m\equiv-n (\mod d)$ and $mn\equiv -1 (\mod d)$ or, $m^2 = 1 (\mod d)$ which need not always have a trivial solution. but from (2) we get $d | [(m-1)(n-1)+(1-m)T_m]$ and $d | mn+1+(mT_m)$ which along with (1) results in $d | (1-m)T_m$ and $d | mT_m$ ; Now it is sufficient to look that $d | T_m$ solves our problem; Since $T_m$ is the fundamental period so $T_m=d ----------(3)$ . Now from ( * ) we also know that $d | m^2-1$ and so $T_m | (m^2-1)$ and thus $T_m=m^2-1$ is the fundamental period

By Euclid's algorithm, $\gcd(a,b) = \gcd(a,ka+b)\text{ where }k\text{ is any integer}$

$\text{take }a=(n+m), b=(mn+1), k=-m,$

$\gcd(n+m,mn+1)=\gcd(n+m,1-m^2)=\gcd(n+m,m^2-1)$

since period $T_m$ is invariant under shifting $n$ to $n-n_0$ ,

we consider period of $g_m(n)=f_m(n-m)=\gcd(n,m^2-1)$

which is obvious that $T_m=m^2-1$ since $m^2-1$ is a constant.

hence $T_m=12^2-1=143$