12 Feb 10
Leonard Euler (1707-1783) was inspired of Pierre de Fermat and his x^n + y^n = z^n where n>2 which is now shown to have no solutions in any n. In the case where n=2 solutions was known by Pythagoras.
Euler thought that neither a^5 + b^5 + c^5 + d^5 = e^5 had a solution. Was he right? (a,b,c,d,e are all integers.) He had no computer. We do. But do we know enough to solve the equation without our silicon buddies?
Question:
Solve the equation a^5 + b^5 + c^5 + d^5 = e^5, where a,b,c,d,e are all integers, if possible.
Was Euler wrong?
12 Feb 10
1 edit
Originally posted by FabianFnas"Solve the equation a^5 + b^5 + c^5 + d^5 = e^5, where a,b,c,d,e are all integers, if possible."
Leonard Euler (1707-1783) was inspired of Pierre de Fermat and his x^n + y^n = z^n where n>2 which is now shown to have no solutions in any n. In the case where n=2 solutions was known by Pythagoras.
Euler thought that neither a^5 + b^5 + c^5 + d^5 = e^5 had a solution. Was he right? (a,b,c,d,e are all integers.) He had no computer. We do. But do we kn ...[text shortened]... a^5 + b^5 + c^5 + d^5 = e^5, where a,b,c,d,e are all integers, if possible.
Was Euler wrong?
a=b=c=d=e=0
Euler also thought that a^4 + b^4 + c^4 = d^4 has no solution in positive integers. This was disproved by Noam Elkies, a famous chess composer.
12 Feb 10
Originally posted by FabianFnasHere are two:
Right. I make a variation of the problem:
Question:
Solve the equation a^5 + b^5 + c^5 + d^5 = e^5, where a,b,c,d,e are all positive integers, if possible.
Not so easy now, eh? 😉
(a, b, c, d, e) = (27, 84, 110, 133, 144)
(a, b, c, d, e) = (55, 3183, 28969, 85282, 85359)
Of course you can multiply all the numbers by any positive integer to get another solution.
Now YOU solve a^5+b^5=c^5+d^5+e^5 (a,b,c,d,e are positive integers).
BTW no non-trivial solution to a^5+b^5=c^5+d^5 or a^5+b^5+c^5=d^5 is known.
12 Feb 10
1 edit
Originally posted by David113You're right in your answer. Are there more solutions?
Here are two:
(a, b, c, d, e) = (27, 84, 110, 133, 144)
(a, b, c, d, e) = (55, 3183, 28969, 85282, 85359)
Now YOU solve a^5+b^5=c^5+d^5+e^5 (a,b,c,d,e are positive integers).
I think I know what algorithm you used to find the answer.
I got my teeth into a^5+b^5=c^5+d^5+e^5. This algoritm is quite complicated, but after some labour I eventually found one answer to be
(a, b, c, d, e) = (14132, 220, 14068, 6237, 5027). Right?
Perhaps I can find more solutions, but not at this late hour.
14 Feb 10
1 edit
Originally posted by FabianFnasI got something different unless you are talking a different order of the numbers. If the sequence is correct that you gave, a=14132, b=220, c=14068, d=6237 and e=5027, I get a^5+b^5+c^5+d^5 = 1.124112117E21 and 5027^5=3.21E18 and change. Am I looking at this wrong?
You're right in your answer. Are there more solutions?
I think I know what algorithm you used to find the answer.
I got my teeth into a^5+b^5=c^5+d^5+e^5. This algoritm is quite complicated, but after some labour I eventually found one answer to be
(a, b, c, d, e) = (14132, 220, 14068, 6237, 5027). Right?
Perhaps I can find more solutions, but not at this late hour.a
I did this stuff on a Casio fx-300ES, If I got more serious I would do it on my rusty trusty HP48 but the results on the casio isn't a matter of the 19th digit wrong, it's almost 3 orders of magnitude off.
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14 Feb 10
Originally posted by sonhouseYou used Fabians solutions for
I got something different unless you are talking a different order of the numbers. If the sequence is correct that you gave, a=14132, b=220, c=14068, d=6237 and e=5027, I get a^5+b^5+c^5+d^5 = 1.124112117E21 and 5027^5=3.21E18 and change. Am I looking at this wrong?
I did this stuff on a Casio fx-300ES, If I got more serious I would do it on my rusty trust ...[text shortened]... ts on the casio isn't a matter of the 19th digit wrong, it's almost 3 orders of magnitude off.
a^5 + b^5 + c^5 +d^5 = e^5
but his solutions are for the equation
a^5 + b^5 = c^5+d^5+e^5...
you just didn't realize they were talking about a different equation than the original. Your casio works fine.
14 Feb 10
2 edits
Originally posted by joe shmomissed that. That would make a tad bit of difference🙂
You used Fabians solutions for
a^5 + b^5 + c^5 +d^5 = e^5
but his solutions are for the equation
a^5 + b^5 = c^5+d^5+e^5...
you just didn't realize they were talking about a different equation than the original. Your casio works fine.
Both = 5.636612043 E20, and I imagine all the rest of the digits line up.
15 Feb 10
1 edit
Originally posted by sonhouseYes, I assure you, the numbers lines up perfectly.
missed that. That would make a tad bit of difference🙂
Both = 5.636612043 E20, and I imagine all the rest of the digits line up.
I used the calculater the Windows system gives me, and it has tremendously many numbers at hand.
Very well, this is the algorithm I used:
I used Googles to find the string "27, 84, 110, 133, 144" and found
http://sites.google.com/site/tpiezas/020
where I fond the original problem, and also David113's supplemental problem, that I 'solved' elegantly.
Check up the link, everyone interesting in mathematics like this!
The original probem, however, I found in a Swedish science magazine dating back to 1992.
15 Feb 10
Originally posted by FabianFnasShouldn't that count as using our silicon buddies? 😛
Yes, I assure you, the numbers lines up perfectly.
I used the calculater the Windows system gives me, and it has tremendously many numbers at hand.
Very well, this is the algorithm I used:
I used Googles to find the string "27, 84, 110, 133, 144" and found
http://sites.google.com/site/tpiezas/020
where I fond the original problem, and also David113 ...[text shortened]...
The original probem, however, I found in a Swedish science magazine dating back to 1992.
15 Feb 10
Originally posted by PalynkaI think you refer to "But do we know enough to solve the equation without our silicon buddies" as I wrote in my opening posting.
Shouldn't that count as using our silicon buddies? 😛
Well, do we? I don't. Does anyone? Do you?
Okay, there are at least two solutions. Are there more? Are there finite numbers of solutions or are there infinitely many?
15 Feb 10
Originally posted by FabianFnasNo, I don't. That's why I asked David how he found out those solutions... I'm also usually stumped by higher order Diophantine equations so was hoping this would help me discover how to approach them better.
I think you refer to "But do we know enough to solve the equation without our silicon buddies" as I wrote in my opening posting.
Well, do we? I don't. Does anyone? Do you?
Okay, there are at least two solutions. Are there more? Are there finite numbers of solutions or are there infinitely many?
PS: He also correctly pointed out that multiplying any solution by a positive integer will also be a solution.
16 Feb 10
Originally posted by PalynkaWouldn't that by itself indicate an infinite number of solutions?
No, I don't. That's why I asked David how he found out those solutions... I'm also usually stumped by higher order Diophantine equations so was hoping this would help me discover how to approach them better.
PS: He also correctly pointed out that multiplying any solution by a positive integer will also be a solution.