However, it breaks down when you have three numbers. As an example, I will use 20, 24, and 32. The product is 15360. The GCD is 4. The LCM is 480. This time, GCD * LCM = 1920. In fact, if abc = LCM(a,b,c) * GCD(a,b,c), it can be proven that a and b are coprime, a and c are coprime, and b and c are coprime. Furthermore, in this case the GCD = 1 and the LCM = abc.
The proof of that theorem requires a new function discovered by me called the multiplicative median. MM(a,b,c) = abc / (GCD(a,b,c) * LCM(a,b,c)). For example, MM(20,24,32) = 15360 / 1920 = 8. By definition, GCD(a,b,c) * LCM(a,b,c) * MM(a,b,c) = abc. It can also be proved that GCD(a,b,c) <= MM(a,b,c) <= LCM(a,b,c). This lemma will be proved below. Now, the above theorem can be proved as follows: abc = LCM(a,b,c) * GCD(a,b,c) = LCM(a,b,c) * GCD(a,b,c) * MM(a,b,c). Thus MM(a,b,c) = 1. Now, 1 divides every positive integer so GCD(a,b,c) >=1 Also, GCD(a,b,c) <= MM(a,b,c) = 1. Thus GCD(a,b,c) = 1. Also, abc = GCD(a,b,c) * LCM(a,b,c) = LCM(a,b,c). Thus LCM(a,b,c) = abc.
Now, let a and b have a common factor greater than 1. Then GCD(a,b) > 1 so LCM(a,b) = ab / GCDa,b < ab. Now, LCM(a,b,c) = LCM(LCM(a,b),c) <= c * LCM(a,b) < c * (ab) = abc. Thus LCM(a,b,c) < abc. However, LCM(a,b,c) = abc so we have a contradiction. Thus a and b are coprime. Similarly, a and c are coprime and so are b and c. Thus the theorem is proved.
I have discovered many more things about this curious MM function. I have thought about when abc = MM(a,b,c)2. Now, MM(a,b,c)2 <= MM(a,b,c) * LCM(a,b,c) <= abc. Thus MM(a,b,c) = LCM(a,b,c) and GCD(a,b,c) = 0. I will spare you the proof but it turns out that a must be xy, b must be xz, and c must be yz for pairwise coprime x, y, and z. However, let abc = MM(a,b,c)3. Now it turns out that given a and b, there is always at least one c such that this is the case. There is no upper limit in how many values are possible. However, for some pairs there is only one c. An example is 324 and 96. c happens to be 559,872. This can be proven to be the only c. Check it. However, some pairs have many. For example, 810,000 and 72,900 have no less than 27 possible values of c! These are listed at the bottom. However, we need bigger and bigger numbers for more values of c. Incidentally, if there is only one possible c, then it follows that c >= ab, with equality if and only if a = b = 1, in which case c = 1. Also, note that a, b, and c in this case cannot be permuted without allowing more solutions. Take a = 96 and b = 559,872. In addition to c being 324, c could also be 165,888; 6,377,292; or 3,265,173,504. In general, if there is only one c for given a, b, there will be 2n values of b for given a, c and 2n values of a for given b, c where c is the only value given a and b. Here n is a positive integer which depends on the values of a and b.
The properties come from the prime factorisation of a, b, and c. For each prime, a has a certain power of that prime, as do b and c. Now, LCM(a,b,c) has the maximum power of those three powers, GCD(a,b,c) has the minimum, and abc has the sum, so MM(a,b,c) has the median of the powers. I chose the name "multiplicative median" because it has the median prime factorisation. As a consequence of this, GCD(a,b,c) <= MM(a,b,c) <= LCM(a,b,c), which proves the lemma from earlier.
If you really want the list of 27 values of c, here it is: 9; 72; 243;
576; 1125; 1944; 6561; 9000; 15,552; 30,375; 52,488; 72,000; 140,625;
243,000; 419,904; 820,125; 1,125,000; 1,944,000; 3,796,875; 6,561,000;
9,000,000; 30,375,000; 52,488,000; 102,515,625; 243,000,000; 820,125,000;
6,561,000,000. If you really, REALLY love crunching numbers and have nothing
else to do, try checking them all to see that if a = 810,000 and b = 72,900,
MM(a,b,c)3 really does equal abc. If you try, you'll figure
out how I generated those 27.
Related entriesComing soon
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