- 12 Feb '10 14:35Leonard 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 14:59 / 1 edit

"Solve the equation a^5 + b^5 + c^5 + d^5 = e^5, where a,b,c,d,e are all integers, if possible."*Originally posted by FabianFnas***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 15:31 / 1 edit

If he didn't mean just positive integers, then if you set e = 0 and a = -b, c = -d (or similar combinations) you have an infinity of solutions...*Originally posted by David113***"Solve the equation a^5 + b^5 + c^5 + d^5 = e^5, where a,b,c,d,e are all integers, if possible."**

a=b=c=d=e=0 - 12 Feb '10 17:27

Here are two:*Originally posted by FabianFnas***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 21:02 / 1 edit

You're right in your answer. Are there more solutions?*Originally posted by David113***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 19:40 / 1 edit

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?*Originally posted by FabianFnas***You're right in your answer. Are there more solutions?**a

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.

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. - 14 Feb '10 19:52

You used Fabians solutions for*Originally posted by sonhouse***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 20:49 / 2 edits

missed that. That would make a tad bit of difference*Originally posted by joe shmo***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 09:17 / 1 edit

Yes, I assure you, the numbers lines up perfectly.*Originally posted by sonhouse***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 10:30

Shouldn't that count as using our silicon buddies?*Originally posted by FabianFnas***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 10:51

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.*Originally posted by Palynka***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 11:02

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.*Originally posted by FabianFnas***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 20:25

Wouldn't that by itself indicate an infinite number of solutions?*Originally posted by Palynka***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.