Was Euler wrong?

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?

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?

"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

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.

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

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...

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? 😉

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? 😉

Here are two:

(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.

Originally posted by David113
(a, b, c, d, e) = (27, 84, 110, 133, 144)
(a, b, c, d, e) = (55, 3183, 28969, 85282, 85359)

How did you find those?

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).

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.

Originally posted by FabianFnas
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 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 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.

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.

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.

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.

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.

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.

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'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.

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.

Shouldn't that count as using our silicon buddies? 😛

Originally posted by Palynka
Shouldn't that count as using our silicon buddies? 😛

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?

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?

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.

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.

Wouldn't that by itself indicate an infinite number of solutions?