As promised, here is the open-ended, rubric-based, holistically scored, performance-assessed, student-constructed first problem from MathNotation's Third Contest:

1. A primitive Pythagorean triple is defined as an ordered triple of positive integers (a,b,c) in which a^{2} + b^{2} = c^{2} and the greatest common factor (divisor) of a, b and c is 1. If (a,b,c) form such a triple, explain why c cannot be an even integer.

Comments

(a) The content here is number theory. Is some of this covered in your district's middle school curriculum or beyond? More importantly, at what point do students begin to formulate and write valid mathematical arguments?

(b) The immediate reaction of most students was that this seemed like a fairly simple problem. However, only a couple of teams scored any points. Perhaps the challenge here was the construction of a deductive argument, although as you will see below, there is one challenging part.

(c) There were two successful approaches used by the teams. Both involved indirect reasoning. Do your students begin to do these in middle school or are "proofs" first introduced in geometry?

(d) I allowed students to assume without proof the following:

(i) The general rules of parity of the sum of two integers

(ii) The square of a positive integer has the same parity as the integer

(e) Interestingly, none of the teams considered an algebraic approach to the one challenging case, i.e., demonstrating that the sum of the squares of two odd integers is not divisible by 4.

If a and b are odd, they can be represented as

a = 2m+1 and b = 2n+1, where m and n are integers.

Then a^{2} + b^{2} = (2m+1)^{2} + (2n+1)^{2} =

(4m^{2} + 4m + 1) + (4n^{2} + 4n + 1) =

4(m^{2} + n^{2}) + 4(m + n) + 2, which leaves a remainder of 2 when divided by 4.

BUT, if c is even, say c = 2k, then c^{2} = 4k^{2}, which is divisible by 4.

(f) The two best solutions came from our first and second place teams, Chiles HS in FL and Hanover Park Middle School in CA. Both used the ideas of congruence modulo 4.

Here is the indirect method used by Chiles:

Let's assume that c can be an even integer. We'll prove by contradiction. An even integer can be summed in two ways:

1. with two even integers or

2. two odd integers

If it is the latter case, then looking at the residuals of modulo 4, the two odd integers summed will be equal to 2, but this is not the case as 2 is not a modulo of 4 residue. If it is the former case, then it does not satisfy the problem as then a, b, and c have common factor of 2. Therefore c must be an odd integer. Q.E.D.

Here is the indirect method used by Hanover Park:

^{2}≡ 0 (mod 4).

We break this into cases based on the parity of a,b.

Case I: Both a and b are even; gcd(a,b,c) ≥ 2 because a,b,c are even, a contradiction.

Case 2: One of a and b is even. Then, a

^{2}+ b

^{2}≡ 0 + 1 ≡ 1

not ≡ 0 (mod 4), a contradiction.

Case 3: Both of a, b are odd. Then a

^{2}+ b

^{2}≡ 1 + 1 ≡ 2

not ≡ 0 (mod 4), a contradiction.

We have covered all cases for a, b with no valid cases. Thus, in a PPT, c cannot be even.

Both of these arguments represent a more sophisticated understanding of mathematics and the methods of proof. Clearly, these students are quite advanced and exceptional, however, I feel many middle school teachers begin early on to encourage their students to explain their thought processes both orally and in writing. Am I right? I would like to hear your thoughts on this...

## 3 comments:

Let's go further with this. Fermat made an observation about positive integers that are the sums of two squares. The proofs are a bit complicated; you can look them up on Wikipedia or Mathworld.

What numbers are sums of two squares (or 1 square--0 counts).

0 = 0²

1 = 1²

2 = 1² + 1²

3: no

4 = 2²

5 = 1² + 2²

6: no

7: no

8 = 2² + 2²

9 = 3²

10 = 1² + 3²

11: no

12: no

13 = 2² + 3²

16 = 4²

17 = 1² + 4²

18 = 3² + 3²

19: no

20 = 2² + 4²

Now, look at the primes in the list. 2, 5, 13, and 17 are sums, while 3, 7, 11, and 19 are not. Do your students notice a distinction?

Now look at the composites. 10 is in the list; its factors 2 and 5 are in also. In fact, if m and n are such sums then their product is. Why? It helps to be familiar with complex numbers.

If m = a² + b², then m = (a+bi) (a-bi).

If n = c² + d², then n = (c+di) (c-di).

mn = (a+bi) (a-bi) (c+di) (c-di)

= (a+bi) (c+di) (a-bi) (c-di)

If (a+bi)(c+di) = (e+fi), then (a-bi)(c-di) = (e-fi), and

mn = (e+fi) (e-fi) = e² + f².

So, we can guess that 2, and odd primes of the form 4n + 1 appear in the list; if that's true then all their products do. Can 7 or 11 appear as a factor of a number in the list? Of course. 7² = 49 appears. And that leads to Fermat's observation:

A positive integer is the sum of two squares if every prime factor with the form 4k+3 appears to an even power.

Eric--

Nice to be talking about math again, number theory in particular.

I've always believed that an understanding of the proof that odd primes are uniquely representable as a sum of squares iff they are of the form 4n+1 is a rite of passage for young mathematicians. Now I'm really curious about Zagier's one-sentence proof of this theorem! Must be a very complex sentence!

Back to the contest...

I specifically chose a fairly straightforward derivation for the test to give all students a chance to at least engage the question. The fact that so many struggled confirms my belief that we should spend a more time early on getting our students to formulate explanations of their reasoning and putting pencil to paper. Writing one's ideas in a coherent manner takes years of practice, never mind the deeper understanding of big mathematical ideas.

I understand. Examinations measure proficiency and understanding. I simply think that examination questions should be related to mathematical history and research. There are short steps from "What squares are sums of two squares?" to "What numbers are sums of two squares?" to "What numbers are sums of three squares?" to "What numbers are sums of four squares?" to Waring's problem.

You might want to discuss with students the distinction between G(n) and g(n) in Waring's problem; similarly, discuss Vinogradov's result on sums of three primes.

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