Cool...!

Long winded way to define $\text{lcm(x,y)gcd(x,y) = xy}.$ We solve for $\text{lcm(52,14) = 2*7*26=364}.$

I have a long solution:

f(52,14)=f(14,52)

=f(14,14+38)

=(52/38) f(14,38)

=(52/38)(38/24) f(14,24)

=(52/24)(24/10) f(14,10)

=(52/10) f(10,14)

=(52/10)(14/4) f(4,10)

=(13*14/10)(10/6) f(4,6)

=(13*14/6)(6/2) f(2,4)

=91*(4/2) f(2,2)

=364

putting y=x in (x+y)f(x,y)=yf(x,x+y), 2x(f(x,x))=x(f(x,2x)) 2xf(x,x)=f(x,2x) 2x=f(2x,x). putting x=2x, y=x, (3x)f(2x,x)=xf(2x,3x) (3)(2x) =f(2x,3x) f(3x,2x)=3(2x) => f(nx,my)=n(mx) 52=26.2 14=7.2 =>f(52,14)=26.7.2=364

It is not hard to see that $f(x,y) = \frac{xy}{{\rm gcd}(x,y)}$ . Since gcd $(52,14) = 2$ , the answer is $52 \cdot 14 / 2 = \fbox{364}$ .

Edit: you could be even more concise: $f(x,y) = {\rm lcm}(x,y)$ . But to prove that property (c) holds for this $f$ , you'd want to use the alternate description I started with, since ${\rm gcd}(x,y) = {\rm gcd}(x,x+y)$ .