Variable location in recursion

You need to wrap your head a little around what the recursion will do: it will call the function recursively to count the pairs from that point on, then (it should) add the pair, if any, found by this instance.

So your function needs to do something with the result of the recursive call, not just return it unchanged. For example, instead of this

elif s[0].lower() == s[1].lower():
    pairs = +1
    return countpairs(s[1:])

you might write this:

elif s[0].lower() == s[1].lower():
    return countpairs(s[1:]) + 1

Something along these lines. You'll need to do a bit more work to get it just right, but I hope you get the idea.

The problems is that the variable pairs resets every recursive call... When using counting recursive algorithms you don't need a variable that counts, that's the beauty Instead, try to think how to recursive call can help you count.

def countpairs(s):
    if len(s)<2:
       return 0                       
    elif s[0].lower() == s[1].lower():      
        return countpairs(s[1:])+1
    else:                                
        return countpairs(s[1:])
print(countpairs('Hello Salaam'))

There you go, in the recursive call the "counter" gets bigger every time it should be, think of the counter as a part of the function stack (or something like that).

You need to fix the syntax: pairs=+1 should be pairs+=1, same for pairs=+0. And you can pass the total into the next level.

def countpairs(s, pairs=0):
    if len(s)<2:
        return pairs                            #base case
    elif s[0].lower() == s[1].lower():       #recursion
        pairs+=1
        return countpairs(s[1:], pairs)
    else:                                   #recursion
        pairs+=0
        return countpairs(s[1:], pairs) 
print(countpairs('Hello Salaam'))  # 2

You can do it by creating a recursive nested function and defining pairs in the outer function. Here's what I mean (with fixes for other issues encountered):

def countpairs(s):
    pairs = 0
    def _countpairs(s):
        nonlocal pairs  # Since it's not local nor global.

        if len(s) < 2:  # base case
            return pairs
        elif s[0].lower() == s[1].lower():
            pairs += 1
            return _countpairs(s[1:])  # recursion
        else:
            return _countpairs(s[1:])  # recursion

    return _countpairs(s)

print(countpairs('Hello Salaam'))  # -> 2

The code will always evaluate to zero because the last recursion will always have the length of s being less than 2. Instead use the global keyword to be able to grab the value of pairs.

numberOfPairs = 0
pairsList = []
def countpairs(s):
    global numberOfPairs
    if len(s)<2:
        print("doing nothing")
        return 0                            #base case
    elif s[0].lower() == s[1].lower():       #recursion
        numberOfPairs+=1
        newString = f"{s[0]} is equal to {s[1]}"
        print(newString)
        pairsList.append(newString)
        return countpairs(s[1:])
    else:
        print(f"nothing happened: {s[0]}")                                   #recursion
        return countpairs(s[1:])

print(f"\nThe returned value of countpairs is: {countpairs('Hello Salaam')}")
print(f"Number of pairs: {numberOfPairs}")
print(pairsList)