tag:blogger.com,1999:blog-8468471599393524678.comments2012-06-08T02:16:18.857-05:00A Suitable Collection of ExamplesRoy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comBlogger56125tag:blogger.com,1999:blog-8468471599393524678.post-89317370268452741452012-05-06T15:17:54.873-05:002012-05-06T15:17:54.873-05:00I'd like to use this example in a talk I'l...I'd like to use this example in a talk I'll be giving at Mathfest in August. Is that ok? If so, how should I credit you. If you want, send me an email at aaron.hill@unt.edufavoritemathproblemshttp://favoritemathproblems.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-29798018565840911452012-02-11T14:03:25.794-06:002012-02-11T14:03:25.794-06:00Three equations, three unknowns, solve however you...<i>Three equations, three unknowns, solve however you like.</i><br /><br />Yeah, that's exactly how the GeoGebra illustration I made does it on the fly. It's a neat algebra problem -- a natural introduction to systems of equations. And the general solution of those equations turns out to be pretty elegant. <br /><br />I've worked out another approach using purely geometrical tools -- i.e. it can be performed on a piece of paper with a compass and straightedge, given three arbitrary points. I wonder if there's a more down-to-earth approach, but I'm becoming convinced that there's not.Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-16663438617708727232012-02-11T13:36:35.483-06:002012-02-11T13:36:35.483-06:00For one, the center of the third circle must be eq...For one, the center of the third circle must be equidistant from the *edges* of the two existing circles. By playing the the geogebra it appears to be a parabolic curve, but I'm unsure how to prove that.<br /><br />As for number two, let the circles by a b c with unknown radii ra rb and rc (I apologize for the lack of proper subscripts). Let the known distances between the centers of the circles be AB BC and AC, where AB is the line segment between the centers of a and b, and so on. Then:<br />AB = ra + rb<br />BC = rb + rc<br />AC = ra + rc<br />Three equations, three unknowns, solve however you like.<br /><br />For number three, find the area of the triangle formed by the centers of the circles. Use the law of (co)sines the determine the interior angles of the triangle, so theta-a is the angle between AB and AC (<180˚). Find the area of circle a inside the triangle by theta-a/pi *ra^2 and likewise for b and c. Add these areas and subtract them from the area of the triangle, and viola.<br /><br />Of course, you've probably figured that out already, and the point is to make the kids think about it - a much harder problem.Maxhttp://www.blogger.com/profile/04207573720797162937noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-42753414305620867232012-01-27T13:59:45.432-06:002012-01-27T13:59:45.432-06:00Here are my calculations: Your center is probably...Here are my calculations: Your center is probably about 30 degrees south. The top of the picture is probably about 40 degrees north. So half your visibility is 70 degrees and your full visibility is 140 degrees. This implies your viewing angle is 40 degrees and so half your viewing angle is 20 degrees. The appropriate right triangle gives: $latex \sin (20) = \frac{4000}{4000 + D}$. This gives $latex D= 7700$ miles.Aaronhttp://www.blogger.com/profile/07091944718087197826noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-71545239275382847192012-01-23T12:12:57.451-06:002012-01-23T12:12:57.451-06:00These are great questions. You can also "go ...These are great questions. You can also "go backwards" in the following way: If you know how far you can see from a certain height, you can compute the radius of the earth. With a little bit of trickery, you can get the earth's rotation to some of the work for you. Here is what you can do:<br /><br />Go to the west side of a large body of water (one you can't see across) just before sunrise. Stand at the waters edge and wait for the first rays of sunlight to come over the horizon. Start counting and put your eyes as close to level with the water as possible (within a few inches is reasonable). Wait for the first rays of sunlight to come over the horizon. Knowing the elapsed time and the height of your eyes (when standing) is enough to calculate the radius of the earth.favoritemathproblemshttp://favoritemathproblems.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-60160697379362339202012-01-10T23:02:33.260-06:002012-01-10T23:02:33.260-06:00Found your little spot on the web and enjoyed this...Found your little spot on the web and enjoyed this entry. I simplified it a good bit and used it for a lesson on exponential functions in my Algebra I class. Students enjoyed it. Thanks for the idea.Brett Edwardshttp://www.brettedwards.comnoreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-50114233153650879562011-12-15T20:09:53.729-06:002011-12-15T20:09:53.729-06:00Thanks for your thoughts, as usual. A couple of no...Thanks for your thoughts, as usual. A couple of notes in response:<br /><br />(1) I would love to take a picture of the lens forming an image of the lamp on a piece of paper (I've been trying), but it is a fairly technically difficult thing to pull off, at least working as I usually do without an extra pair of hands. <br /><br />I think a more important Dan Meyer message that's applicable here is to let students form their own questions. Specifically, if this were actually directed at students in a classroom, I think the ideal thing would be to stop somewhere before the moment that I wrote "For example, what would happen if..." and let students lead the way more. As far as I can see, this is one situation where it would be pretty difficult to come up with a "wrong question." <br /><br />In fact, I wouldn't be too surprised if, perhaps given more extensive measurements showing the horizontal and vertical asymptotes more clearly, and a bit of time (probably more than I generally have), the students were able to come up with the thin lens formula themselves, albeit in a less elegant form.<br /><br />(2) I don't use a Mac, and very few of my students do, while GeoGebra is really universally available. But aside from all that, it's also sort of nice to have an immediate legitimate reason to solve the equation for one of its variables (though certainly there are other good reasons you might want to do so anyway). Sometimes technological limitations have nice pedagogical side-effects.Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-64896987161931763602011-12-15T13:35:34.690-06:002011-12-15T13:35:34.690-06:00Fun stuff. Three quick thoughts.
(1) Can't hel...Fun stuff. Three quick thoughts.<br />(1) Can't help but see this through the Dan Meyer lens. I want a photo of your lamp/lens/image situation rather than a line drawing. It really would be more visually compelling.<br />(2) Grapher, which is buried in the Applications/Utilities folder of every Mac can do implicitly defined functions, including in three variables.<br />(3) Love the 1/p+1/q=1/f formula, as it connects to my common-numerator fraction addition algorithm. Basically, 1/p+1/q=1/(pq/(p+q)). That is, the numerator stays the same and the denominator becomes a fraction that is the product over the sum of the original denominators. Thus, your f is the product over the sum of p and q, or 400/50=8. And that 8 holds throughout (roughly).christopherdanielsonhttp://christopherdanielson.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-52441603161818186932011-11-27T20:29:54.073-06:002011-11-27T20:29:54.073-06:00I can't remember for the life of me where I go...I can't remember for the life of me where I got this set of data, but I do know that it is authentic, and as I recall, the sample was chosen somewhat randomly.Roy Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-26083590662847267392011-09-09T15:00:17.407-05:002011-09-09T15:00:17.407-05:00That much is true.That much is true.R. Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-70127474740448889892011-09-09T13:17:12.011-05:002011-09-09T13:17:12.011-05:00I think what you're searching for here is the ...I think what you're searching for here is the notion of "density" as it pertains to a subset of the natural numbers (see, for example, http://en.wikipedia.org/wiki/Natural_density). Indeed, multiples of three can be put into a 1-1 correspondence with the natural numbers, but if we look on the interval [1,N] for any N, and assign the discrete uniform distribution to counting numbers on this interval, we have roughly a 1/3 probability of a randomly chosen natural number being divisible by 3 (as you've pointed out). Letting N goes to infinity gives rise to the notion of density, which in my understanding is the proper terminology, though I wouldn't be surprised if more ambiguous words like "proportion" show up in the literature.<br /><br />There are other notions of density that one can define on the natural numbers, too. It is quite a rich and interesting topic!Matthttp://www.mathgoespop.comnoreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-62873282655559536312011-09-08T16:29:06.364-05:002011-09-08T16:29:06.364-05:00Good point. However, one can find the phrase "...Good point. However, one can find the phrase "proportion of all integers" used in scholarly papers here and there (e.g. <a href="http://bit.ly/pNhFlY" rel="nofollow">http://bit.ly/pNhFlY</a>), so maybe there are some precedents for this very problematic idea.<br /><br />In the end, your point is too strong to ignore. I'd rephrase along the lines of "If <i>N</i> is very large, what proportion of the counting numbers up to <i>N</i> are multiples of 4, 6, or 9?"<br /><br />Thanks for keeping me honest, so to speak.R. Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-40279933266697719362011-09-08T13:16:18.644-05:002011-09-08T13:16:18.644-05:00If it's all counting numbers, I can put the mu...If it's <i>all</i> counting numbers, I can put the multiples of 3 in 1:1 correspondence with the non-multiples. So what is the meaning of <i>proportion</i> when we're dealing with infinite sets?christopherdanielsonhttp://christopherdanielson.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-50589172759015035162011-09-06T16:05:34.925-05:002011-09-06T16:05:34.925-05:00I used a similar problem in my discrete math class...I used a similar problem in my discrete math class recently: How many of the counting numbers from 1 to 1,000,000 are multiples of 2, 3, or 5? That's slightly messier. Here I mean all counting numbers.<br /><br />I was kind of hoping this might raise some philosophical issues. I wondered a little bit myself about the implications of asking a question like this about all counting numbers, but I think it's reasonable. After all, can't we say that 1/3 of all counting numbers are multiples of 3, for example? Any extra insights you might have would be appreciated.R. Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-54555937581559022482011-09-06T08:02:17.656-05:002011-09-06T08:02:17.656-05:00Question, Mr. Wright...do you mean all counting nu...Question, Mr. Wright...do you mean <i>all</i> counting numbers? Or do you mean <i>any suitably large subset</i> of counting numbers?<br />And does it matter?christopherdanielsonhttp://christopherdanielson.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-58666146456265289812011-08-11T12:12:37.371-05:002011-08-11T12:12:37.371-05:00Ok, so I can finally put this proof to rest! Third...Ok, so I can finally put this proof to rest! Third time's the charm. :)<br /><br />m^p + n^p = 2^q<br />For non-negative integers m, n, p and q show that m=n=2^R, where R is a non-negative integer<br /><br />let m = (2^r)*M and n = (2^s)*N<br />(so that M and N contain the odd part of m and n)<br /><br />Case 1: r =/= s (assume s > r, wlog)<br />(2^pr)*M^p + (2^ps)*N^p = 2^q<br />M^p + [2^p(s-r)]*N^p = 2^(q-pr)<br /><br />or odd + even = even, and we have a contradiction!<br /><br />Case 2: r = s<br />(2^pr)*M^p + (2^pr)*N^p = 2^q<br />M^p + N^p = 2^(q-pr)<br /><br />Subcase i) M = N = 1<br />1 + 1 = 2^(q-pr)<br />which is true for q = pr + 1<br />i.e. 2^pr + 2^pr = 2^(pr + 1)<br /><br />Subcase ii) M =/= 1, M=N<br />M^p + M^p = 2^(q-pr)<br />M^p = 2^(q-pr-1)<br /><br />or odd = even, contradiction.<br /><br />Subcase iii) M =/= 1, M =/= N (assume M > N, wlog)<br />Let t = M - N, t is even<br /><br />Subsubcase a) p is even<br /><br />(t + N)^p + N^p = 2^(q-pr)<br />2*N^p + p*t*[N^(p-1)] + summation{k=0,p-2}(p<b>C</b>k)*[t^(p-k)]*N^k = 2^(q-pr)<br /><br />where we have used binomial expansion, but pulled out the last two terms from the summation.<br /><br />N^p + (p/2)*t*[N^(p-1)] + summation{k=0,p-2}(p<b>C</b>k)*{[t^(p-k)]/2}*N^k = 2^(q-pr - 1)<br /><br />N^p is odd<br />(p/2)*t*[N^(p-1)] is even<br />every term in the summation is even, since t is even, and always appears in degree > 1.<br /><br />putting it all together, we have odd + even + summation of even terms = even, and another contradiction.<br /><br />Subsubcase b) p is odd<br /><br />(M + N)(M^p + N^p) = (M + N)*2^(q-pr)<br />M^(p+1) + N^(p+1) + MN*[M^(p-1) + N^(p-1)] = (M + N)*2^(q-pr)<br />(t + N)^(p+1) + N^(p+1) + MN*[M^(p-1) + N^(p-1)] = (M + N)*2^(q-pr)<br />2*N^(p+1) + (p+1)*t*N^p + MN*[M^(p-1) + N^(p-1)] + summation{k=0,p-1}[(p+1)<b>C</b>k]*[t^(p-k+1)]*N^k = (M + N)*2^(q-pr)<br /><br />where we have again used binomial expansion for the p+1 expansion, but pulled out the last two terms from the summation.<br /><br />N^(p+1) + [(p+1)/2]*t*N^p + MN*{[M^(p-1) + N^(p-1)]/2} + summation{k=0,p'-2}(p'<b>C</b>k)*[t^(p'-k)]*N^k = (M + N)*2^(q-pr-1)<br /><br />And, we have odd + even + even + summation of even terms = even, which is a contradiction. <br />Their is an exception for the case of p=1, where the third term,<br />MN*{[M^(p-1) + N^(p-1)]/2} = MN*[(M^0 + N^0)/2] = MN which is still odd.<br /><br />I was working on showing a necessary relationship between M and N for p=1, but am satisfied and will probably leave it!<br /><br />So m = n = 2^Q (with the exception of cases where p=1)mrnchowhttp://mrnchow.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-91030456167467314582011-08-10T22:53:25.433-05:002011-08-10T22:53:25.433-05:00(Near the end I wrote the term "(p/2)*t*[N^(p...(Near the end I wrote the term "(p/2)*t*[N^(p1)]" twice. They should read "(p/2)*t*[N^(p-1)]")mrnchowhttp://mrnchow.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-13085913260663840392011-08-10T22:49:30.000-05:002011-08-10T22:49:30.000-05:00Hey R. Wright,
That deadline was for me, so that ...Hey R. Wright,<br /><br />That deadline was for me, so that I would work!<br />I got stuck yesterday proving it for odd starting powers (I'd be interested in seeing a proof for the power 3, actually). I'm not sure my original proof actually generalizes!<br /><br />I didn't have any new insights on it today, so I'll just post the proof for even starting powers. :)<br /><br />---<br /><br />m^p + n^p = 2^q [1]<br />For non-negative integers m, n, p, and q show that for [1] to be true, m = n = 2^Q<br /><br />Let m = (2^r)*M and n = (2^s)*N<br />(so that M and N contain the odd part of m and n)<br /><br />Case 1: r =/= s (assume s > r, wlog)<br />(2^pr)*M^p + (2^ps)*N^s = 2^q<br />M^p + [2^p(s-r)]*N^p = 2^(q-pr)<br /><br />or odd + even = even, and we have a contradiction!<br /><br />Case 2: r = s<br />(2^pr)*M^p + (2^pr)*N^s = 2^q<br />M^p + N^p = 2^(q-pr) [2]<br /><br />Subcase i) M = N = 1<br />1 + 1 = 2^(q-pr)<br />which is true for q = pr + 1<br />i.e. 2^pr + 2^pr = 2^(pr + 1)<br /><br />Subcase ii) M =/= 1 (assume M > N, wlog)<br />Let t = M - N, t is even<br /><br />Now,<br />M^p + N^p<br />= (t + N)^p + N^p<br />= 2*N^p + p*t*[N^(p-1)] + summation{k=0,p-2}(pCk)*[t^(p-k)]*N^k [3]<br /><br />where we have used binomial expansion, but pulled out the last two terms from the summation. note that this is exactly where a starting power of 1 breaks down, as the summation is from k=0 to p-2. (also note that I would have used LaTeX code, if I wasn't so rusty... because I don't like my ad-hoc convention here!)<br /><br />Now divide [3] by 2,<br />N^p + (p/2)*t*[N^(p1)] + summation{k=0,p-2}(pCk)*{[t^(p-k)]/2}*N^k<br /><br />N^p is odd<br />for even p, (p/2)*t*[N^(p1)] is even<br />every term in the summation is even, since t is even, and always appears in degree > 1.<br /><br />returning to [2] then, we have odd + even + summation of even terms = even, and we have another contradiction.<br /><br />NB: if t = (2^u)*T where u > 1, this holds contradiction holds for odd p, as well. My gut feels that it holds for all odd p, but the full proof has eluded my attempts!<br /><br />So for even p, m = n = 2^Q<br /><br />---<br /><br />I'll continue to work at odd p in the back of my head, but other work calls right now. Let me know your thoughts!mrnchowhttp://mrnchow.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-7219901140890180722011-08-10T00:47:24.899-05:002011-08-10T00:47:24.899-05:00I'll post a better proof within a day!
Hey, n...<i>I'll post a better proof within a day!</i><br /><br />Hey, no pressure! I'm more than happy with what you've already given.R. Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-45630135349328261292011-08-08T22:11:15.045-05:002011-08-08T22:11:15.045-05:00Hey R. Wright,
Thanks for your questions (and the...Hey R. Wright,<br /><br />Thanks for your questions (and the original post)! It has been enlightening to think about this easy to understand problem. In particular, your questions have led me to a conceptual easier (and cleaner) way of understanding it (and why my previous proof is slightly wrong)!<br /><br /><i>for x to be a positive integer, only the log_2(2^2j) term can survive from n^2, by definition of prime factors.</i><br /><br />What I'm stating here is that for one term, log_2(R^2r), this cannot be integer for any base R (except R=2). This is true because for R^2r to be represented in 2^y, it must have factors of 2 in it, which is impossible due to the prime factorization. I'm also making a stronger claim that any combination of log_2(R^2r) + log_2(S^2s) terms (and more than 2) cannot be integer (and hence x will not be integer). This can be seen by returning them to a log of a product: log_2[(R^2r)*(S^2s)], and noting that again, that it contains no factors of 2, and thus can't be represented as some 2^y (where y is an integer).<br /><br />But as @jamestanton quickly pointed out to me, this is the complicated way of doing it (http://bit.ly/n7fzC4). It's far easier and conceptually cleaner to put n into the form (2^r)*N, as opposed to prime factorizing it. Now we know that N is odd (N^2 is also odd), and 2^r contains the even part of n, if n is even.<br /><br />Now,<br />log_2(n^2) = log_2[(2^2r)*N^2] = log_2(2^2r) + log_2(N^2) = 2r + 2log_2(N)<br />but N is odd, and for log_2(N) to be integer, it must be 1.<br /><br /><i>Where would the proof break down (as it must) for a starting power of 1?</i><br /><br />This is an awesome question! It shows that I did things slightly out of order with this proof. It's technically wrong when I said <i>m/2^j has to be integer, for x to have a chance at being integer, which eliminates all prime factors of m other than the prime factor of 2 (e.g. k',l',... = 0).</i> An obvious example is m=6 *sheepish*<br /><br />I got ahead of myself, and should have first said <i>for log_2[1 + (m/2^j)^2] to be integer, 1 + (m/2^j)^2 has to be at least even (i.e. (m/2^j)^2 has to be odd).</i><br /><br />To prove this, we can say m=(2^s)*M so that,<br />log_2[1 + (m/2^j)^2] = log_2{1 + [(2^s-j)*M]^2} and [(2^s-j)*M]^2 has to be odd. 2^s-j contains the even part, and has to equal to 1 (i.e. s=j) since an even number times an odd number is always even (and an even number squared is even).<br /><br />Now we have to show when 1 + M^2 can be put into 2^y form. we know that M is odd, so let M = L + 1, where L is even.<br /><br />1 + M^2 = 1 + L^2 + 2L + 1 = L^2 + 2(L + 1)<br /><br />Let's presume that L^2 + 2(L + 1) = 2^y, where y is a non-negative integer<br /><br />(L^2)/2 + (L + 1) = 2^(y-1)<br /><br />since L is even, (L^2)/2 has to be even, L + 1 is odd, and 2^(y-1) is even. This statement can never be true, except for when L = 0, hence M = 1 (and m=2^s-j).<br /><br />This step is also the exact point where the starting power of 1 breaks down. <br /><br />For 1 + M = 2^y or L + 2 = 2^y, there is no restriction possible, since we now have L + 2, not 2L + 2. We need an extra L to factor out the 2 and make the even and odd argument.<br /><br />I think can do this with the prime factorization way too… but I anticipate it will be long and ugly!<br /><br />---<br /><br />I now see the statement "Two square numbers add to a power of two. What can you say about the two squares? Two cubes also add to a power of two. So ..?" as a deep statement about the nature of odd and even numbers. Because things are represented in the base 2, it has deep implications to the even and oddness of those squares.<br /><br />To be honest, I feel my proof is badly structured and a round-about way of doing things. My instinct to use prime factorization was a bit off (though it did lend to the even and oddness of the problem, since 2 is the only even prime), and I'd like to clean it up. I'll post a better proof within a day!<br /><br />(the fact that this statement is also a statement about even and oddness leads me to believe that James has a visual way of doing this, and I'd like to see it!)nathanhttp://www.blogger.com/profile/12308783582149308839noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-51339914149501370672011-08-08T12:38:27.316-05:002011-08-08T12:38:27.316-05:00This claim,
If the sum of two squares (or cubes) ...This claim,<br /><br /><i>If the sum of two squares (or cubes) is a power of 2, then the squares (or cubes) are equal powers of 2 </i><br /><br />turns out to be very instructive. It is very straightforward — just about anyone, whatever their mathematical background, could understand what is being claimed — but so far no one has come up with a simple proof. In that, it is somewhat akin to Fermat's Last Theorem. Also, as I said, its converse is pretty obviously true, unlike the converse of Fermat's Last Theorem.R. Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-90050193469048381522011-08-08T12:31:09.096-05:002011-08-08T12:31:09.096-05:00Thanks for your comment, and sorry for the delay i...Thanks for your comment, and sorry for the delay in responding. One thing in your proof I'm not sure I follow: <br /><br /><i>for x to be a positive integer, only the log_2(2^2j) term can survive from n^2, by definition of prime factors.</i><br /><br />Could you explain this line? That is, why must log_2(3^2k) and so on be zero?<br /><br />Also, as you say,<br /><br /><i>this works for any starting power (i.e. squares, cubes, etc...)</i><br /><br />Where would the proof break down (as it must) for a starting power of 1?R. Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-42921557483736730122011-08-05T00:25:28.695-05:002011-08-05T00:25:28.695-05:00I'm sure someone with a stronger math backgrou...I'm sure someone with a stronger math background can clean this up (alas I'm a physicist, and not trained in rigorous proofs!)<br /><br />Start with,<br /><br />2^x = n^2 + m^2<br />x = log_2(n^2 + m^2)<br /><br />prime factorize n and m, so that<br /><br />x = log_2{[(1^i)(2^j)(3^k)(5^l)...]^2 + [(1^i')(2^j')(3^k')(5^l')...]^2}<br />where i,i',j,j'... are non-negative integers<br />(we can drop 1^i and 1^i' w.l.o.g.)<br /><br />So,<br /><br />x = log_2[(2^2j)(3^2k)(5^2l)... + (2^2j')(3^2k')(5^2l')...]<br />x = log_2{(2^2j)[(3^2k)(5^2l)... + (2^2j'-2j)(3^2k')(5^2l')...]}<br /><br />and factor each prime factor of n^2 off of m^2 accordingly<br /><br />Then,<br /><br />x = log_2{(2^2j)(3^2k)(5^2l)...[1 + (2^2j'-2j)(3^2k'-2k)(5^2l'-2l)...]}<br /><br />using log of a product,<br /><br />x = log_2(2^2j) + log_2(3^2k) + log_2(5^2l) + ... + log_2[1 + (2^2j'-2j)(3^2k'-2k)(5^2l'-2l)...]<br /><br />for x to be a positive integer, only the log_2(2^2j) term can survive from n^2, by definition of prime factors. thus k,l,... = 0<br /><br />So,<br /><br />x = log_2(2^2j) + log_2[1 + (2^2j'-2j)(3^2k')(5^2l')...]<br /><br />return m^2 to its composite form<br /><br />x = log_2(2^2j) + log_2[1 + (m^2)/(2^2j)]<br />x = log_2(2^2j) + log_2[1 + (m/2^j)^2]<br /><br />m/2^j has to be integer, for x to have a chance at being integer, which eliminates all prime factors of m other than the prime factor of 2 (e.g. k',l',... = 0).<br /><br />Now,<br /><br />x = log_2(2^2j) + log_2[1 + (2^j'/2^j)^2]<br /><br />for log_2[1 + (2^j'/2^j)^2] to be integer, 1 + (2^j'/2^j)^2 has to be at least even (i.e. (2^j'/2^j)^2 has to be odd). This can only happen when j' = j. Thus n = m.<br /><br />So, for integer x, 2^x = n^2 + m^2 only when n = m = 2^j where j is a non-negative integer<br /><br />For completion,<br /><br />x = log_2(2^2j) + log_2(1 + 1^2) = 2j + 1<br /><br />which is the 'intuitive' result of 2^x = n^2 + n^2 = (2^1)(n^2), where n = 2^j<br /><br />----<br /><br />NB: this works for any starting power (i.e. squares, cubes, etc...)<br /><br />QED?nathanchowhttp://mrnchow.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-16149326034012301582011-07-23T01:19:48.481-05:002011-07-23T01:19:48.481-05:00Do you mean that I should have made the illustrati...Do you mean that I should have made the illustration(s) to scale? I exaggerated the relative dimensions to make the geometric principles clearer.<br /><br />But that made me realize that GeoGebra would be a really simple way to do some calculations. When I draw everything to scale, it turns out that from the ISS, the Earth's radius should appear to be only 5.97 times the thickness of the atmosphere (by the Kármán line).<br /><br />Conversely, if the apparent thickness of the orange layer is 1/72 the radius of the Earth, then the orange layer is only 10 km up, according to my geometric calculations, which matches pretty darn well with the location of the tropopause...R. Wrighthttp://www.blogger.com/profile/11151362439746234042noreply@blogger.comtag:blogger.com,1999:blog-8468471599393524678.post-70456347955124598012011-07-22T23:21:58.592-05:002011-07-22T23:21:58.592-05:00A more useful example would have the earth (radius...A more useful example would have the earth (radius 6368km), the atmosphere (100km), and the ISS (402km about the surface) all to scale using known values and run the math. That gives the ideal answer from the height of the ISS, which we can compare with the accepted value (63.7:1).<br /><br />I think the radius of the earth is known from Newton's law of gravity and the Kármán line is slightly arbitrarily defined to be 100km, since the atmosphere fades out smoothly and you have to draw a line somewhere. So the real values are know without looking at images from space after all.Max Goldsteinhttp://dethorningSTEM.comnoreply@blogger.com