Update:
(1) See the visualization for the difference of squares posted on 1-9-08.
(2) Read the comments in this post for considerable clarification and instructor guidelines and suggestions. Mathmom's and Eric's comments are particularly insightful.
This post can be developed into an activity for prealgebra through first-year algebra students (or even 2nd year algebra). The last part is more challenging.
The focus here is on developing a method/strategy that can be used to solve similar Diophantine equations. The other objective is to introduce the ideas and methods of proof. This problem may later be used to solve a recent math contest problem for which I obtained permission to discuss on this blog. I am fully aware that many students will 'solve' these equations by Guess-Test methods, but they need to go further.
STUDENT/READER PROBLEMS/ACTIVITIES
(a) Prove there is only one solution in positive integers for the equation:
M2 - N2 = 12
Note: If we omit the word positive, what would the solution(s) be?
(b) Determine all positive integer solutions:
M2 - N2 = 15
(c) Determine all positive integer solutions:
M2 - N2 = 36
(d) Let's investigate for what positive integer values of P, M2 - N2 = P has NO solutions in positive integers.
(i) Determine at least 5 positive integer values of P for which the above equation has no positive integer solutions.
(ii) (More challenging) Describe all values of P for which the above equation has no solutions. Justify your result.
Note: All students should have success with (i), although some may struggle to find 5 values. Part(ii) should challenge the student who has finished the other parts in rapid order and sits there complacently!
Additional Comment: If P is itself a perfect square, our equation is obviously related to the most famous equation in geometry. Thus, if P = 9 or P = 16, for example, students should recognize something! For this reason you may want to have students consider these values when doing this investigation. More to come...
Tuesday, January 8, 2008
M^2 - N^2 = 12...Prove there is only one solution in positive integers and much more
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8 comments:
I can tell you for sure none of my pre-algebra students can "prove" any of this. They might be able to guess/check to get solutions, but I'm skeptical of even that.
Diophantine equations were college algebra for me, and I remember nothing. I can see that there is only one answer to (a) by "brute force" (looking at the increasing pattern of differences between the squares) but I see no easy way to prove it; certainly nothing accessible to pre-algebra students, even if they were comfortable with the notion of a "proof" (which they are not). So, am I missing something, or is this not really accessible to a pre-algebra level.
By the way, what was the contest?
With such problems, it's always best to find a multiplicative formulation. For example, the Goldbach conjecture asks whether every even integer ≥ 4 is the sum of two primes. The answer is unknown. But every positive integer is the product of primes in exactly one way.
So, write (a) as:
12 = M²−N² = (M+N) (M−N).
But the two factors differ by 2N; so they are both even or both odd. Since the product is even, they're both even. Factorizing 12 gives
M+N = 6
M−N = 2.
M = 4, N = 2.
(b) By the same argument,
M+N = 15, M−N = 1, or
M+N = 5, M−N = 3.
M=8, N=7, or
M=4, N=1.
(c) Again:
M+N = 18, M−N = 2, or
M+N = 6, M−N = 6.
M = 10, N = 8, or
M = 6, N = 0.
(d) What sort of number is a square? Do squares have special characteristics? I do doubt whether any high schooler would realize without prompting that squares ≡ 0 or 1 (mod 4). If they do, then they can find that M²−N² ≡ −1, 0, or 1 (mod 4).
So, numbers ≡2 (mod 4) satisfy d.
Going to higher mathematics, one of the first ways to factorize numbers came from this equation. It was invented by Blaise Pascal. I'll write again about it later.
Mathmom, Eric--
This problem is certainly ambitious for the younger student. Eric's in-depth analysis is certainly beyond the expectations of middle schoolers, however, take a look at today's post which uses an area diagram to demonstrate the factorizaton of the difference of squares. This is appropriate for prealgebra students IMO.
Certainly, most students would begin by guessing values to solve these equations. The instructor should have them organize their findings in a table (see below) from which conjectures could be more easily made. This might be more than enough for the prealgebra students. The justifications would be for the older student -- I should have made that clearer.
P............Solutions
1............None
2............None
3............2^2-1^2
4............None
5............3^2-2^2
6............None
7............4^2-3^2
etc.
Eric, in addition to P=1, P=2 and P=4, which are exceptional cases, I believe some students would be able to see the generalization that non-representable values are of the form 2⋅(an odd integer) =
2(2n+1) = 4n+2, which is equivalent to your modular representation.
Here's why:
P = (M+N(M-N)
Case I:If P is odd (and greater than 1), we CAN find a representation as follows:
M+N=P
M-N=1
Therefore, M= (P+1)/2 and N = (P-1)/2 would be a solution.
Case II: The problem arises when P is even. If P is divisible by 4 (but not equal to 4), then we can find two DIFFERENT EVEN factors of P and proceed as before. If P is even but NOT divisible by 4, we CANNOT find a pair of factors that are BOTH EVEN, so there is no solution. This is precisely the case that P equals 2 times an odd integer or of the form 4N+2 as you suggested.
Note that 2 itself is of the form 2 times an odd!
Does this make sense or am I missing something? I believe some students would discover this if they proceed in a systematic fashion.
Mathmom--
I was referring to the last question from the high school math league question from the December contest. I will post it soon.
also, Eric, Mathmom--
The restriction to positive integer solutions was critical since it excluded 1 and 4 from being representable.
High school students should consider the significance of using perfect square values for P and its connection to Pythagorean Triples.
(d):
Obvoiusly M^2 - N^2 = P has a soution if
P is odd or a square number itself (then
let N be 0 and M is sqrt(P))
If P is even there can only be a solution
for M and M if P contains square factors.
Example:
P = q^2 * P'
M^2 - N^2 = q^2 * P'
this leads to
(M/q)^2 - (N^q)^2 = P'
Recursivly this equation only has a
solution for the properties described above.
The values of M and N can then be obtained by back
substitution. And we can conclude that M^2 - N^2 = P
has no solution if P is even or square factor free.
Dude--
How about 18? 18 is not square-factor free but I don't believe M^2 - N^2 = 18 has a solution.
Also, I restricted M,N to be positive so M^2 - N^2 = 4 has no solution either! Read my comment above in which I described the even values of P for which there is no solution. Someone needs to verify this!
Another way to approach this is to consider all numbers of the form n(n+e) where e is any even number. If any number satisfies this equation, then the two squares can be easily found. Take, for instance, 5(5+6) = 55. Divide e by two, in this case 6/2 = 3, square the result, and add it to the original number. Thus we have: 55 + (6/2)^2 = 55 + 9 = 64, then 64 - 9 = 55. Hence if you know the factors of n, you can determine whether or not n can be expressed as the difference of two squares.
Take 18 as another example. There is no way to arrange the factors of 18 such that n(n+e) = 18: 18=6*3, 18=9*2, 18=18*1. Therefore 18 cannot be expressed as the difference of two squares. Incidentally this shows that all odd numbers have at least one such expression (including primes) -- o*1 = o (example: 7*1 = 7). So 1(1+6) = 7 thus 7 + 9 = 16 and 16 - 9 = 7. That also follows from the figurate representation of squares:
16 = 1 + 3 + 5 + 7
9 = 1 + 3 + 5
Therefore 16 - 9 = 7, and it is trivial to see that all odd numbers can be expressed as the difference of successive squares.
Andrew--
Interesting analysis...
The idea of completing the square is inspired. In effect, a number can be represented as a difference of squares if one can add a square to the number and produce another square. I followed through your argument algebraically:
n(n+e) n^2 + en.
Adding (e/2)^2, the standard term when completing the square, produces
n^2 + en + (e/2)^2 = (n+e/2)^2.
Thus, n(n+e) = (n+e/2)^2 - (e/2)^2.
Applying this approach to odd integers of the form 2k+1 produces:
2k+1 = (2k+1)((2k+1)+(-2k)) =
((2k+1)+(-k))^2 - (-k)^2 =
(k+1)^2 - k^2, a nice way to represent any odd number! For example, 11 = 2x5 + 1 = (5+1)^2 - (5^2) or 11 = 6^2-5^2.
The only numbers that are NOT representable as a difference of squares are those of the form 2*odd, i.e., even numbers NOT divisible by 4.
Even numbers that are MULTIPLES OF 4 are easily shown to be representable using your approach:
This fits into your form since 4k = 2(2k) = 2(2 + (2k-2)), so n = 2 and e = 2k-2. Thus, 4k = (2+(k-1))^2 - (k-1)^2 = (k+1)^2 - (k-1)^2.
For example, 24 = 4x6 = (6+1)^2 - (6-1)^2, i.e., 24 = 7^2 - 5^2. Cool...
18 of course is not divisible by 4, therefore, it is not representable.
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