8 people in 4 teams with different pairs in each team each day for 7 days without repeated pairs or anyone being in the same within 3 days Ok I am a Scout Leader and on our 7 day summer camp we have 8 Leaders and will have the Scouts in 4 different patrols or teams.
I want to set up a rota for the Leaders so that they can be assigned to help the Scout Patrols for a day in pairs.
I want all the Leaders to work with every other Leader for a day which I know is possible as there are 28 different pairings with 8 people and helpfully there are 7 days and 4 patrols making 28 spaces to allocate to the 28 pairs.
I also want to make sure each Leader helps with each Patrol at least once, again this is not too hard to achieve (its basically an 8 team round robin tournament at 4 grounds) but the stumbling block I am having is that ideally I do not want any Leader to help with the same Patrol without at least a 2 day gap (e.g. help Monday and not back there until Thursday). I am not sure this is possible but I certainly do not want any Leader helping with the same patrol on consecutive days if i can help it.
I do have a Maths degree but its well over 10 years since I completed it now and I do not work in Mathematics so I am unfortunately very rusty on this kind of thing but I am sure there must be a logical way of solving this rather than just trial and error.
Sorry if I have posted in the wrong place or not made my question clear enough
 A: I have two flaws: The seventh day involves a repetition, and C doesn't supervise Patrol 1.  

FG EH DB AC
    AH BG CF DE
    BE CD AG HF
    FD AE HB GC
    GH FB EC AD
    AB CH DG EF
    HD GE AF BC

I have seen two different ways to organize the leaders.  The first is Ed's, in which pairs such as A and B play a special role:  In round two, A pairs C and B pairs D, but in round 3 they swap, so A pairs D and B pairs C.  The seven days are: 
$$\begin{array}{cccc}AB&CD&EF&GH\end{array}$$
$$\color{blue}{
\begin{array}{cccc}AC&BD&EG&FH\\
AD&BD&EH&FG\end{array}}$$
$$\color{green}{\begin{array}{cccc}AE&BF&CG&DH\\
AF&BE&CH&DG\end{array}}$$
$$\color{red}{\begin{array}{cccc}AG&BH&CE&DF\\
AH&BG&CF&DE\end{array}}$$
During three consecutive days, a patrol must have six different leaders.  It turns out that these days must be either two of one colour and one of another colour; or the black day and one each of two colours.  The possible day order is then either

black, red, green,green,blue,blue,red; or
  green,red,green,black,blue,red,blue  

or a similar arrangement with different colours.  However, when leaders are assigned patrols for the first three days, it becomes impossible to assign them for the fourth day.
So this arrangement of leaders does not lead to a solution.  
Another arrangement of leaders uses a 2x4 rectangle.  A stays in the $1,1$ spot, and the other leaders cycle through the other spots.  Each column gives a pairing.  The arrangement is this:
$$\begin{array}{cccc}AH&BG&CF&DE\\
AG&HF&BE&CD\\
AF&GE&HD&BC\\
AE&FD&GC&HB\\
AD&EC&FB&GH\\
AC&DB&EH&FG\\
AB&CH&DG&EF\end{array}$$
Three consecutive days can only give each patrol six different leaders if two of the days are consecutive rows, and the other is either two rows later or two rows earlier.  So, for example, R1,R3,R4,R6,R7,R2 is possible, but R5 cannot be fitted in.
So this arrangement of leaders does not give a solution.
A: This is the Social Golfer Problem. 
day 1:  12 34 56 78
day 2:  57 68 13 24
day 3:  46 17 28 35
day 4:  38 25 47 16
day 5:  14 67 23 58
day 6:  26 48 15 37
day 7:  45 36 27 18  
Edit -- I've made some changes for the other constraint.  
A: Do you mean something like this?
AB CD EF GH
CE FG AH BD
DG EH BC AF
BF DA EG CH
AC GF DH BE
DE BH CF AG 
BG DF AE HC
A: This is my best attempt, mainly by trial and error
AB  CD  EF  GH
FH  AE  CG  BD
CE  GB  DH  AF
AD  CH  BF  EG
EH  DF  AG  CB
GF  AC  BH  ED
BE  GD  CF  AH
I think I have every pair and everyone visits each patrol and I have at least avoided anyone remaining at the same patrol 2 days running.
I would still love to know if their is a more mathematical way to get a solution
