Year: 1975
Topic: Algebra Number Theory

Problem:
In the additive group of ordered pairs of integers $(m,n)$ [with addition defined componentwise: $(m,n)+(m',n')=(m+m',n+n')$] consider the subgroup $H$ generated by the three elements \[(3,8), \quad (4,-1), \quad (5,4).\] Then $H$ has another set of generators of the form \[(1,b), \quad (0,a),\] for some integers $a,b$ with $a>0$. Find $a$.

Solution:
Also one must have $b\equiv 5 \pmod{7}$. \textbf{Proof:} The subgroup $H$ must contain $4(3,8)-3(4,-1)=(0,35), \ 4(5,4)-5(4,-1)=(0,21)$, and then $2(0,21)-(0,35)=(0,7)$. Now $(0,7)$ and $(1,b)$ will generate $H$ iff $(1,b)$ is in $H$ and there exist integers $u,v,$ and $w$ such that \[(3,8)=3(1,b)+u(0,7), \quad (4,-1)=4(1,b)+v(0,7), \quad (5,4)=5(1,b)+w(0,7).\] These hold iff $8=3b+7u, \ -1=4b+7v, \ \text{and } 4=5b+7w.$ With $b=5+7k, \ k \text{ any integer, the desired coefficients } u,v, \text{ and } w \text{ exist in the form } u=-1-3k, \ v=-3-4k, \ w=-3-5k.$ It now suffices to let $k=0$ and to note that $(1,5)=(4,-1)-(3,8)+2(0,7) \text{ is in } H.$ The answer is $a=\boxed{7}$.