tag:blogger.com,1999:blog-8231784566931768362.post6195912604187664904..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: M^2 - N^2 = 12...Prove there is only one solution in positive integers and much moreDave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-8231784566931768362.post-21897811338502403632008-01-24T11:16:00.000-05:002008-01-24T11:16:00.000-05:00Andrew--Interesting analysis...The idea of complet...Andrew--<BR/>Interesting analysis...<BR/>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:<BR/>n(n+e) n^2 + en.<BR/>Adding (e/2)^2, the standard term when completing the square, produces<BR/>n^2 + en + (e/2)^2 = (n+e/2)^2.<BR/>Thus, n(n+e) = (n+e/2)^2 - (e/2)^2.<BR/><BR/>Applying this approach to odd integers of the form 2k+1 produces:<BR/>2k+1 = (2k+1)((2k+1)+(-2k)) =<BR/>((2k+1)+(-k))^2 - (-k)^2 =<BR/>(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.<BR/><BR/>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. <BR/><BR/>Even numbers that are MULTIPLES OF 4 are easily shown to be representable using your approach:<BR/>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. <BR/>For example, 24 = 4x6 = (6+1)^2 - (6-1)^2, i.e., 24 = 7^2 - 5^2. Cool...<BR/><BR/>18 of course is not divisible by 4, therefore, it is not representable.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-70342234565772123322008-01-24T09:57:00.000-05:002008-01-24T09:57:00.000-05:00Another way to approach this is to consider all nu...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.<BR/><BR/>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:<BR/><BR/>16 = 1 + 3 + 5 + 7<BR/>9 = 1 + 3 + 5<BR/><BR/>Therefore 16 - 9 = 7, and it is trivial to see that all odd numbers can be expressed as the difference of successive squares.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-20104946705934953772008-01-10T13:25:00.000-05:002008-01-10T13:25:00.000-05:00Dude--How about 18? 18 is not square-factor free b...Dude--<BR/>How about 18? 18 is not square-factor free but I don't believe M^2 - N^2 = 18 has a solution. <BR/>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!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-52193254531829925532008-01-10T13:00:00.000-05:002008-01-10T13:00:00.000-05:00(d):Obvoiusly M^2 - N^2 = P has a soution ifP is o...(d):<BR/><BR/>Obvoiusly M^2 - N^2 = P has a soution if<BR/>P is odd or a square number itself (then<BR/>let N be 0 and M is sqrt(P))<BR/><BR/>If P is even there can only be a solution<BR/>for M and M if P contains square factors.<BR/><BR/>Example:<BR/><BR/>P = q^2 * P'<BR/>M^2 - N^2 = q^2 * P'<BR/><BR/>this leads to<BR/><BR/>(M/q)^2 - (N^q)^2 = P'<BR/><BR/>Recursivly this equation only has a<BR/>solution for the properties described above.<BR/><BR/>The values of M and N can then be obtained by back<BR/>substitution. And we can conclude that M^2 - N^2 = P<BR/>has no solution if P is even or square factor free.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-47086549749053186202008-01-09T07:08:00.000-05:002008-01-09T07:08:00.000-05:00Mathmom--I was referring to the last question from...Mathmom--<BR/>I was referring to the last question from the high school math league question from the December contest. I will post it soon.<BR/><BR/>also, Eric, Mathmom--<BR/>The restriction to positive integer solutions was critical since it excluded 1 and 4 from being representable.<BR/><BR/>High school students should consider the significance of using perfect square values for P and its connection to Pythagorean Triples.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-75927892178950993832008-01-09T06:36:00.000-05:002008-01-09T06:36:00.000-05:00Mathmom, Eric--This problem is certainly ambitious...Mathmom, Eric--<BR/>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.<BR/>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.<BR/>P............Solutions<BR/>1............None<BR/>2............None<BR/>3............2^2-1^2<BR/>4............None<BR/>5............3^2-2^2<BR/>6............None<BR/>7............4^2-3^2<BR/>etc.<BR/><BR/>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) =<BR/>2(2n+1) = 4n+2, which is equivalent to your modular representation.<BR/><BR/>Here's why:<BR/>P = (M+N(M-N)<BR/><BR/>Case I:If P is odd (and greater than 1), we CAN find a representation as follows:<BR/>M+N=P<BR/>M-N=1<BR/>Therefore, M= (P+1)/2 and N = (P-1)/2 would be a solution.<BR/><BR/>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.<BR/>Note that 2 itself is of the form 2 times an odd!<BR/><BR/>Does this make sense or am I missing something? I believe some students would discover this if they proceed in a systematic fashion.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-84282762096294476632008-01-09T00:41:00.000-05:002008-01-09T00:41:00.000-05:00With such problems, it's always best to find a mul...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.<BR/><BR/>So, write (a) as:<BR/><BR/>12 = M²−N² = (M+N) (M−N).<BR/><BR/>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<BR/><BR/>M+N = 6<BR/>M−N = 2.<BR/><BR/>M = 4, N = 2.<BR/><BR/>(b) By the same argument, <BR/><BR/>M+N = 15, M−N = 1, or<BR/>M+N = 5, M−N = 3.<BR/><BR/>M=8, N=7, or<BR/>M=4, N=1.<BR/><BR/>(c) Again:<BR/><BR/>M+N = 18, M−N = 2, or<BR/>M+N = 6, M−N = 6.<BR/><BR/>M = 10, N = 8, or<BR/>M = 6, N = 0.<BR/><BR/>(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).<BR/><BR/>So, numbers ≡2 (mod 4) satisfy d.<BR/><BR/>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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-28054520275374761052008-01-08T21:53:00.000-05:002008-01-08T21:53:00.000-05:00I can tell you for sure none of my pre-algebra stu...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.<BR/><BR/>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. <BR/><BR/>By the way, what was the contest?mathmomhttps://www.blogger.com/profile/05869925405540832241noreply@blogger.com