text: Its equivalent to proving\nm*n = (lcm(m,n) ) * ( gcd(m,n) )\n\nNow, i'll have to use basic number theory. The idea is to show that the number of times a prime factor of m*n occurs on the LHS is equal to the number of times a prime factor of (lcm(m,n) ) * ( gcd(m,n) ) occurs on the RHS.\nNote : LHS = Left Hand Side\n       RHS = Right Hand Side\n\nFor any prime factor p of m*n, call it p. Now say p occurs to the power of h in m and power of k in n. That is to say m "contains" p^h while n contains p^k. Note that h,k are nonnegative integers. \n\nNow, in lcm(m,n), we would choose p^ [ min(h,k) ] while\nin gcd(m,n) we would choose  p^ [ max(h,k) ]   \nNote that min (h,k) means minimum of h and k, similarly  max (h,k)means maximum of h and k . For example, if h=2 and k=3, then min (2,3) = 2. while max(2,3) = 3. If say h = k, then obviously in both lcm and gcd of m*n we choose p^h to be in its product. This is true for EVERY prime factor of m*n.\n\nSo, consider the product (lcm(m,n) ) * ( gcd(m,n) ).\nFor EACH prime factor p of m*n, we have shown above that it occurs as { p^ [ min(h,k) ] } * { p^ [ max(h,k) ]   }, that is our RHS product contains \n  p^ [ min(h,k) +  max(h,k) ] = p^(h+k)\nNow, in our LHS, p occurs as p^h * p^k = p^(h+k) too!!!\nThus, we have shown for each prime factor, p of m*n, it occurs as the same product in both the LHS and RHS. And thus\nwe have LHS = RHS. Done.\n\nOne question may circle in your mind, that is how i know that [ min(h,k) +  max(h,k) ] = (h+k)\nWell, this is a fact.\n\nFurther, if you want to show it, consider 3 cases.\n\nCase 1. ( h = k)\nHere min(h,k) = min(h,h) = h\nand  max(h,k) = max(h,h) = h\nSo, [ min(h,k) +  max(h,k) ] = h + h = 2h.\nAnd (h+k)= h + h = 2h.\n\nCase 2. ( h < k)\nHere min(h,k) =  h, because h is smaller.\nand  max(h,k) = k, because k is bigger.\nSo, [ min(h,k) +  max(h,k) ] = h + k \n\nCase 3. ( h > k )\nHere min(h,k) =  k, because k is smaller.\nand  max(h,k) = h, because h is bigger.\nSo, [ min(h,k) +  max(h,k) ] = k + h