It’s All About “Prod-Difs”—Vive la Différence and Multiply
Pick any four positive integers. For example, 5, 14, 17, and 23 give a collection of four positive integers. Now, form the differences of pairs of these numbers by subtracting the smaller number from the larger in each pair (although this won’t really matter). In our case, this process would give us the following:

(14 – 5) = 9                            (17 – 14) = 3

(17 – 5) = 12                          (23 – 14) = 9

(23 – 5) = 18           and         (23 – 17) = 6

Now, form the product of all these differences. This gives us 9 x 12 x 18 x 3 x 9 x 6 = 314,928, in our case. Let’s call the result the Prod-Dif of the four integers. So, the Prod-Dif of the collection {5, 14, 17, 23} is 314,928. We can form the Prod-Dif of any collection of four positive integers. Then we have a very interesting problem to ponder:

What is the largest integer that divides evenly into all Prod-Difs? That is, what is the greatest common factor of the collection of all Prod-Difs? And of course, WHY?  

Some Ponderings: 

Recall that a Prod Dif of four integers a, b, c, and d is the product of all the pair-wise differences of those integers. So we could find the Prod Dif of a, b, c, and d as follows:

Prod Dif (a, b, c, d) = (a – b) x (a – c) x (a – d) x (b – c) x (b – d) x (c – d).

For example, the Prod Dif (2, 4, 11, 15) = (2 – 4) x (2 – 11) x (2 – 15) x (4 – 11) x (4 – 15) x (11 – 15) = (-2) x (-9) x (-13) x (-7) x (-11) x (-4) = 72072.  

And the Prod Dif (1, 2, 3, 4) = (1 – 2) x (1 – 3) x (1 – 4) x (2 – 3) x (2 – 4) x (3 – 4) = (-1) x (-2) x (-3) x (-1) x (-2) x (-1) = 12.

Note that the Prod Dif (2, 5, 6, 6) = 0, since the Prod Dif of four numbers where one is repeated will have a factor of zero when we form the pair-wise differences.

Our problem to ponder was to find the largest integer that divides evenly into all Prod Difs. Clearly 1 divides all Prod Difs, but the question is, is there a number greater than 1 that divides all Prod Difs? We noted that we might as well form all the pair-wise differences so that the factors are positive, because the sign of the final product won’t matter for a divisor.

One conjecture that students are likely to make as they generate a list of Prod Difs is that all Prod Difs are even numbers—and, therefore, that 2 will divide all Prod Difs. Also, they are likely to realize fairly soon that the smallest nonzero Prod Dif that is possible is Prod Dif (1, 2, 3, 4) = 12, as shown above. This may lead students to conjecture that in fact 12 is itself the largest number that divides evenly into all Prod Difs, since it works for the Prod Dif (1, 2, 3, 4) as well as other Prod Difs that they try.) If we look at the parity of the four numbers (that is, whether they are odd or even), the possibilities are that all four numbers are odd numbers, that three are odd and 1 is even, that two are odd and two are even, that one is odd and three are even, or that all four are even numbers. Because Odd – Odd = Even, and Even – Even = Even, when we form the pair-wise differences of the four numbers, we will always get at least two pair-wise differences that yield an even number. So in fact, there will be two factors of 2 in every Prod Dif, or 2 x 2 = 4 will divide every Prod Dif. In order to prove that 12 will in fact divide every Prod Dif, we need to find a way to convince ourselves that 3 will also always be a factor of any Prod-Dif. Because we can’t find any counterexamples when we start building Prod Difs, we become pretty certain this is the case, but an absence of evidence is not evidence of absence! We must find a way to show that 3 does divide every Prod Dif. Ponder away, and tune in next time for further reasoning about why 3 also divides every Prod Dif.

Further Ponderings:

In our ponderings last month, we saw that there will always be two factors of 2 in every Prod Dif, so that 2 x 2 = 4 will divide every Prod Dif. To finish an argument that 12 will divide every Prod Dif (remember: we weren’t able to find a counter example; 12 always worked!), we need to find a way to convince ourselves that 3 will also always be a factor of any Prod-Dif.  Recall how Prod-Difs are formed. We pick any four integers, (a, b, c, d), form all possible pairs of differences (that is, (a – b), (a – c), (a – d), (b – c), (b – d), and (c – d)), and then form the product of those 6 numbers:

PROD DIF (a, b, c, d) = (a – b) x (a – c) x (a – d) x (b – c) x (b – d) x (c – d).

Why does at least one of these differences have to be divisible by 3?

Consider that for every integer, there are three possibilities:

it is a multiple of 3 to start with (that is, it has the form 3n for n some integer),   

or it is 1 more than a multiple of 3 (it has the form 3n + 1),

or it is 2 more than a multiple of 3 (it has the form 3n + 2).

(Once we get to “3 more than a multiple of 3,” we’re just back in the multiples of 3!)

Here’s another way to think about this: whichever integer you pick out is going to belong to one of these three sets:

{ … –12, –9, –6, –3, 0, 3, 6,  9,  12 ...},

or {…–11, –8, –5, –2, 1, 4, 7, 10, 13 …},

or {…–10, –7, –4, –1, 2, 5, 8, 11, 14 …}.   

These three subsets cover all the integers, and every integer belongs to exactly one of the subsets (called a partition).

In forming Prod-Difs, we picked 4 integers. So, at least two of them have to come from the same one of these three sets—we are “pigeonholed,” so to speak. So 3 must also be a factor of every Prod-Dif.

Now, for a nice extension—what if we picked five integers, and made a Prod-Dif? What then? What would the greatest common factor of all Prod-Difs of 5 integers be? How about 6? And… Well, you get the picture.