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Old 02-13-2009   #11 (permalink)
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Arrow Re: One question answered

Quote:
Originally Posted by Tirtle
is this on the right track?

Binary Quadratic Forms as Equal Sums of Like Powers
Quote:
Originally Posted by Titus Piezas III
Fauquembergue, 1898:
(4x^4-y^4)^4 + (4x^3y)^4 + (4x^3y)^4 + (2xy^3)^4 + (2xy^3)^4 = (4x^4+y^4)^4
Quote:
Originally Posted by CraigD View Post
Maybe. There are a lot of tracks in sight. Which ones are right … well, that’s the rub.
I know the routine.

48^4 + 64^4+64^4+32^4+32^4=80^4

So the expression for powers of 4 that I quoted is missing a term it seems, namely another (2xy^3)^4. It should read: (4x^4-y^4)^4 + (4x^3y)^4 + (4x^3y)^4 + (2xy^3)^4 + (2xy^3)^4+(2xy^3)^4 = (4x^4+y^4)^4

My solution is for x=2 & y=2.


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semantics is not always just pedantic quibbling. ~ douglas r. hofstadter
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Old 02-13-2009   #12 (permalink)
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Arrow Re: A Generalization of Fermat’s Last Theorem

for x=1 & y=1,

(4x^4-y^4)^4 + (4x^3y)^4 + (4x^3y)^4 + (2xy^3)^4 + (2xy^3)^4+(2xy^3)^4 =
 (4x^4+y^4)^4 =

2^4+2^4+3^4+4^4+4^4=5^4

How very Pythagorean-ish.


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semantics is not always just pedantic quibbling. ~ douglas r. hofstadter
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Old 02-13-2009   #13 (permalink)
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Arrow Re: A Generalization of Fermat’s Last Theorem

Quote:
Originally Posted by Turtle View Post
Substituting in this equation u = 2 & v = 2 we get a negative term in the sum:

832^5-768^5+320^5+1600^5+2368^5 =2432^5

Edit additions:
u=1 & v=2
40^5+800^5+2399^5-10399^5 +10401^5=2401^5

u=2 & v=1
7^5+43^5+57^5+80^5+100^5 =107^5


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semantics is not always just pedantic quibbling. ~ douglas r. hofstadter

Last edited by Turtle; 02-14-2009 at 09:13 AM..
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Old 02-21-2009   #14 (permalink)
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Arrow Re: A Generalization of Fermat’s Last Theorem

Quote:
Originally Posted by Turtle View Post
Substituting in this equation u = 2 & v = 2 we get a negative term in the sum:

832^5-768^5+320^5+1600^5+2368^5 =2432^5

Edit additions:
u=1 & v=2
40^5+800^5+2399^5-10399^5 +10401^5=2401^5

u=2 & v=1
7^5+43^5+57^5+80^5+100^5 =107^5
For powers of 5 Craig gives this example: 11^5+9^5+7^5+6^5+5^5+4^5=12^5

The equation I found from Sastry in 1934, (u^5+25v^5)^5 + (u^5-25v^5)^5 + (10u^3v^2)^5 + (50uv^4)^5 + (-u^5+75v^5)^5 = (u^5+75v^5)^5 does not generate this example using integers for u & v. I wonder if there exist rational numbers u & v in Sastry's equation that generate Craig's example?

I tried working out u = 2 & v = .5, but I was using the Windows scientific calculator & I think it was clipping off the decimal result past 23 places because the equation was very very close to balanced. Is there a balanced equation here or not?

So thens, beyond powers of 5 there is no known equation for powers of 6. Dare we go down that road? Is it time to add a 3rd variable perhaps??

Dat's dat.


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