I. Two years ago today my family left my childhood home of 16+ years and moved 40 minutes farther north, from Snellville to Buford. It's interesting, but as I've mentioned earlier, I never had much trouble thinking of our new house as "home," and even memorizing our new phone number (which conveniently has some sort of a mathematical pattern :-D) and address wasn't an issue. But today, while at a local store purchasing shoes *groan*, I reverted back to my old address when asked for my zip code. Ironic, since today was the two year anniversary of our move.

II. One year ago today I posted my first blog entry. Since then I've posted 216 entries (including this one), which is neat since 216 is 6^3, and 6 is the first perfect number. Also 216 is the product of 8 and 27, and 8 and 27 are the first two (nontrivial) perfect cubes. And no I didn't plan that :-D. But in all serious, I'm thankful for this past year of blogging, the many friends the Lord has sent my way, and the many,

*many*lessons He has taught me. I don't know just how long this season of blogging will continue, but I'm enjoying it while it is upon me :). Thank you to all my blogging friends for being such an encouragement and challenge to me this past year!

## 32 comments:

Happy Birthday to you

Happy Birthday to you

Happy Birthday dear Blog

Happy Birthday to you

Isn't 216 so beautiful?

In honour of your anniversaries, here's a math problem:

a^4 + b^4 + c^4 + d^4 + e^4 + f^4 = 200007, where a..f are integers.

Find all solutions.

*sigh* I'm getting nowhere here. I've had all sorts of ideas as to how to approach the problem, but unfortunately none of them have panned out :(. Anyone else (Att: Adrian *ahem*) want to take a stab at it? Brother Dear doesn't currently have access to his cell phone, or I'd make him write a computer program.

I will try working on it some more, and maybe something will come to light :).

Brute force wins. In speed, albeit not in elegance. 44 million combinations later, I have the solution...but I won't post it for the benefit of anyone who might attempt the puzzle. :-D

I prefer the phrase "proof by exhaustion."

So here's our problem for John:

a^5 + b^5 + c^5 + d^5 + e^5 + f^5 = 38,808,088 where a,. . .,f are integers.

It seems like I never have time anymore to comment on your posts (though I still do eventually read most of them thanks to the handy-dandy Google Reader...good recommendation!)... but I just wanted to say a "Happy First Birthday!" to you and your blog! I've very much enjoyed many of the posts you've had over the year (or a little less since I think I came across your blog in late fall last year)...keep up the good work!

Happy Blog-aversary! I've really enjoyed your blog. I'm so glad I found it!

Happy birthday! I can't believe it's been a year since we started sewing curtains together! :-) I've enjoyed your blog very much!

You know, if you had stayed in Snellville you'd be closer to me! I think. Perhaps not. But traffic is much easier getting to Buford so perhaps it's good you moved.

Oh dear, I'm starting to feel like I made a mistake. It

seemedlike a water-tight and elegant proof...If you mean the proof that there

isno solution, Brother Dear thinks he has proved it. Perhaps he can post it :).Whew! Yes, there is no solution. Yeah, tell him to go for it.

*backpedal*

First, yes, after writing a computer program for me, Ben confirmed that there are no solutions. I had suspected as much, and didn't want to waste hours searching for a solution that did not exist :). That's what brothers are for.

Second, no, he didn't write a proof. He did, but it contained mistakes, so we have no valid proof :(. I misunderstood him before. He thinks a complete proof with no errors would be extremely difficult, possibly even a proof by exhaustion :-P.

Third, I think the proof lies with the unit digit, since fourth powers can only end in a 0, 1, 5, or 6, and 0 is only rarely. If I had the time or care, I'd maybe try to work a proof, but I think the best proof would probably use modular arithmetic, which isn't quite as fresh on my mind as it was a few years ago.

That would be the trouble, then.

OK. Consider this:

p^4 ≡ 0 or 1 (mod 8) for all p.

That is, all powers of 4 give a remainder of 0 or 1 when divided by 8. If p is even, p^4 = (2n)^4 = 16n^4, which is divisible by 8.

If p is odd, p^4 = (2n+1)^4 = 16n^4 + 32n^3 + 24n^2 + 8n + 1, which always has a remainder of 1 when divided by 8.

Hence, in a^4 + b^4 + c^4 + d^4 + e^4 + f^4 = 200007, the

maximumremainder of the left hand side of the equation is 6, but the remainder of the right hand side is always 7, when both sides are divided by 8.Hence the equation has no integer solutions.

Ah, very clever. Modular arithmetic is a beautiful thing.

Beautiful, hmm...

(I guess beauty IS in the eye of the beholder! ;-))

Hey, no fair! I didn't even see the problems before eager-beavers John and Ben attacked it. Well, anyway.

No one seems to have tackled the powers of five problem yet, so I thought I'd take that one from John. I wrote a C++ program to solve it, and found the following: using 772971436 different combinations, the closest I got was

6^5 + 6^5 + 20^5 + 26^5 + 29^5

= 38808077,

which is unfortunately 11 away from the correct value. This leads me to posit that there is no solution.

A number-theoretic proof is currently eluding me, though I'm working on it.

In Christ.

Yes, for the powers of five equation, modular arithmetic doesn't work.

Ya'll are nuts. (I meant that in the best way, of course.)

Um, Adrian, this post is 2 days old. I'd hardly call Ben and John "eager beavers." You just need to check blogs more frequently :).

For anyone who is interested, Brother Dear's girlfriend Stephanie came up with the same proof as John; she just didn't post it in time.

Father Dear is so happy that people are actually trying his problem. He was the one who wanted me to post the powers of five problem. I'm trying to decide if I should take pity on you two or not. Was this a cruel problem? Nah. Perhaps your program neglected all the details, though, Adrian? :-D

I'm surrounded by math and computer geeks. I feel close to normal. . .

Hmm. I'll keep on thinking about it.

For what it's worth, 38,808,088 = 2 x 2 x 2 x 11 x 11 x 47 x 853

Wait a second: negative integers make a difference, since the odd powers don't cancel the signs out. Time to go back to the drawing board.

I found a solution! After altering my C++ program to check the following domain for all the integers a through f: -32 <= number <= 32, my program displayed the following solution:

(-11)^5 + 5^5 + 17^5 + 22^5 + 26^5 + 29^5 = 38808088.

Brute force wins again.

And very sly comment about neglecting details in my program.

In Christ.

*applauds for Adrian* ;)

Well done!

Is that solution unique?

Come to think about it, with negative numbers, choosing ± 32 as the parameters becomes quite arbitrary.

You know, it's very hard working out powers of five in one's head...

I have no idea if that solution is unique. Any clue, Adrian?

The reason that I said before

I'm trying to decide if I should take pity on you two or not. Was this a cruel problem? Nah.was because as far as I know, there is no nice-and-neat way to solve that problem. We just took six random integers and raised them to the fifth. I didn't think anyone would actually try to solve it. I was just trying to show how quick it is to generate one of those problems, regardless of how long it takes to solve one. But then someone *ahem* actually did solve it - with C++. That issonerdy, by the way. *rolls eyes* I would never do that. . .. . . I'd just have my brother do it for me ;).

Up to a permutation of the integers (the problem clearly exhibits that kind of symmetry), I really don't know. I suppose I might get the gumption to modify my program slightly (It wouldn't take much) to sift through looking for more solutions. You know, one thought did occur to me: if one of the integers is large enough, then the difference between that integer raised the fifth and the next biggest integer raised to the fifth would be larger than your target value. You could use that fact to limit the breadth of your search.

Yes, it's nerdy, especially since it only took me about an hour to do it. ;-)] Probably would take Ben half that time, eh?

In Christ.

I was just trying to show how quick it is to generate one of those problemsYes, I thought so. ;)

if one of the integers is large enough, then the difference between that integer raised the fifth and the next biggest integer raised to the fifth would be larger than your target value. You could use that fact to limit the breadth of your search.No, you couldn't. Because the sum of the

other fourpowers might be negative.There might well be an inifinity of solutions.

Okay, well bonus challenge. One of you prove that there are infinite solutions, or prove that there is only one :). . . Assuming, that is, that one of the above is true. We could have a finite number of multiple solutions. Or you can let the problem rest and save your time :).

I smiled at your comment about working 5th powers in your head, John. That's something I would do. Mental arithmetic fascinates me. In fact, I started your problem by working all the fourth powers without a calculator, but I decided my time would be better served by utilizing one.

I think it's funny that you spent an hour writing a program for one problem, Adrian. I have no idea how long it would take Brother Dear, actually. Likely less time, but C++ is a mystery to me. I can do italics, bold, and web links with HTML. That's the closest I can come to computer programming :-D.

Hey! I took two semesters of C++ in college. I got excellent. Then I forgot it all. Oh well. I guess these math problems are beyond my solving...

That was my contribution to this ah, interesting discussion.

You're right, John. I should change it thus: Assume one of the integers

athroughfis the largest in magnitude. This makes sense, because it would be pointless to cancel that term out with its negative to get zero (that would imply you're performing this sum with only four integers raised to the fifth). There must be a smallest integer such that its fifth power is more than 38,808,088 greater than five times the sum of the next smallest integer raised to the fifth. You could use that as your smallest integer. How's that sound, John?So you're interested in the smallest integer

xsuch thatx^5 - 38,808,088 > 5(x-1)^5.That integer appears to be around -23, according to Mathematica.

I decline, however, to delve more deeply, as I imagine it would take more number theory than I know. Sorry, Susan. :-)]

In Christ.

Belated Happy Blogging Anniversary, Susan! I'm just catching up on all my tons of blogs after being out of town.

I enjoy your blog...except when I don't understand all the math stuff (as in all these comments on this post!. :)

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