A question about Fermat's Last Theorem

[SLD]

Fermat's last theorem famously states that there are no integers that satisfy the equation:

xⁿ+yⁿ=zⁿ

For n>2

But what about for this equation?

xⁿ+yⁿ+zⁿ=wⁿ

Or adding even further variables?

SLD

 

[lpetrich]

 Euler's sum of powers conjecture generalizes Fermat's last theorem:
\( \sum_{k=1}^m (a_k)^n = b^n \)
is only possible for integer a's and b if m >= n where n > 2.

For n = 3, we get the cube case of Fermat's last theorem, while for n = 4 and 5, we have counterexamples. Here are the smallest ones:

95800^4 + 217519^4 + 414560^4 = 422481^4
27^5 + 84^5 + 110^5 + 133^5 = 144^5

The first one was found in 1988, and the second one in 1966 on a CDC 6600 computer.

I verified the latter one by brute force on my iMac. Its CPU is a 2.3-GHz Intel Core i5 chip. Its optimizations were compiler optimizations, generating semi-increasing sequences directly, and precalculating all the power values. I used C++ STL binary_search to check on whether a sum value was also a power value. I checked up to 150, checking through 22 million sets of values, and it took 0.4 seconds to do so. Trying up to 300 required 7 seconds, checking through 344 million possibilities, and returned the above result and that result multiplied by 2.

 

[SLD]

 Euler's sum of powers conjecture generalizes Fermat's last theorem:
\( \sum_{k=1}^m (a_k)^n = b^n \)
is only possible for integer a's and b if m >= n where n > 2.

For n = 3, we get the cube case of Fermat's last theorem, while for n = 4 and 5, we have counterexamples. Here are the smallest ones:

95800^4 + 217519^4 + 414560^4 = 422481^4
27^5 + 84^5 + 110^5 + 133^5 = 144^5

The first one was found in 1988, and the second one in 1966 on a CDC 6600 computer.

I verified the latter one by brute force on my iMac. Its CPU is a 2.3-GHz Intel Core i5 chip. Its optimizations were compiler optimizations, generating semi-increasing sequences directly, and precalculating all the power values. I used C++ STL binary_search to check on whether a sum value was also a power value. I checked up to 150, checking through 22 million sets of values, and it took 0.4 seconds to do so. Trying up to 300 required 7 seconds, checking through 344 million possibilities, and returned the above result and that result multiplied by 2.

Ahh. Thanks. Disproven through brute computer force.

 

[lpetrich]

 CDC 6600 -- that 1966 computer -- about 1000 times slower than my iMac, and a heck of a lot larger.

I did it in C++ because I needed the speed -- I didn't want a lot of object-management overhead. I was pleasantly surprised at how fast it was. I had an earlier version where I took an inverse power with the cmath pow() function, rounded it off, then took a power of it to compare with the original sum. It was not quite as fast, about two or three times slower.

 

[steve_bank]

Multiplication and divisions take the longest time to perform digitally.

If you have a repetitive function with fixed constants it can be a lot faster to recalculate

multiplications and divisions into fixed real numbers.

Depending on the compeer the order in which you set up a series of calculations like an equation can affect sped, and accuracy.

In the 80s I got y first computer to learn numerical methods. I started out converting FORTRAN in the texts to C. There was a book in the day Numerical Recopies In C.

Not all software uses the hardware accelerator in the CPU chip. You sometimes have to specify it in the compiler. If the an app uses software multiplication instead of hardware it can be slow.

On an 8Mhz PC of the day a 4k Fast Fourier Transforms could take 30 minutes. Adding the hardware math coprocessor would bring it down to tens of seconds.

 

[Bomb#20]

Fermat's last theorem famously states that there are no integers that satisfy the equation:

xⁿ+yⁿ=zⁿ

For n>2

But what about for this equation?

xⁿ+yⁿ+zⁿ=wⁿ

Or adding even further variables?

SLD

3³+4³+5³=6³

 

[Kharakov]

Fermat's last theorem famously states that there are no integers that satisfy the equation:

xⁿ+yⁿ=zⁿ

For n>2

But what about for this equation?

xⁿ+yⁿ+zⁿ=wⁿ

Or adding even further variables?

SLD

3³+4³+5³=6³

What is the smallest consecutive 4 4th powers that add up to a 4th power?

Smallest consecutive 5 5th powers that add up to a 5th?

....

What is the general formula for the gap between their first (lowest) powers?

 

[lpetrich]

I'll list some smallest examples:

(power, how many to add)

(4,3): 95800^4 + 217519^4 + 414560^4 = 422481^4
(5,4): 27^5 + 84^5 + 110^5 + 133^5 = 144^5

(3,3): 3^3 + 4^3 + 5^3 = 6^3
(4,4): 30^4 + 120^4 + 272^4 + 315^4 = 353^4
(5,5): 19^5 + 43^5 + 46^5 + 47^5 + 67^5 = 72^5
(6,6): (none known; it values have lower limit 730,000)
(7,7): 127^7 + 258^7 + 266^7 + 413^7 + 430^7 + 439^7 + 525^7 = 568^7
(8,8): 90^8 + 223^8 + 478^8 + 524^8 + 748^8 + 1088^8 + 1190^8 + 1324^8 = 1409^8
(9,9): (none known)

 Lander, Parkin, and Selfridge conjecture
\( \sum_{k=1}^m (a_k)^p = \sum_{k=1}^n (b_k)^p \)
abbreviated as (p, m, n). For odd p, it reduces to the Euler conjecture, while for even p, it doesn't.

Some simple cases:
(4,2,2): 59^4 + 158^4 = 133^4 + 134^4
(6,3,3): 3^6 + 19^6 + 22^6 = 10^6 + 15^6 + 23^6

 

[lpetrich]

Here is something related to a special case of the Lander, Parkin, and Selfridge conjecture.

 1729: Mathematician G.H. Hardy had once visited Srinivasa Ramanujan when the latter was hospitalized:

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."

1729 = 1^3 + 12^3 = 9^3 + 10^3
Allowing negative integers gives
9^1 = 6^3 + (−5)^3 = 4^3 + 3^3

The nth  Taxicab number is the smallest number that can be expressed as the sum of the cubes of two positive integers in n different ways: Tc.

Six of them are known, and there are an infinite number of them (A011541 - OEIS). The next one is
87539319 = 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3

The  Generalized taxicab number is the smallest number that can be written as the sum of power p of m positive integers in n different ways: Tc(p,m,n):

Tc(1,2,2) = 4 = 1 + 3 = 2 + 2
Tc(2,2,2) = 50 = 1^2 + 7^2 = 5^2 + 5^2
Tc(3,2,2) = 1729 = 1^3 + 12^3 = 9^3 = 10^3
Tc(4,2,2) = 635318657 = 59^4 + 158^4 = 133^4 + 134^4

It is not know if there are any numbers Tc(p,2,n) with p >= 5.

 

[lpetrich]

Not all software uses the hardware accelerator in the CPU chip. You sometimes have to specify it in the compiler. If the an app uses software multiplication instead of hardware it can be slow.

On an 8Mhz PC of the day a 4k Fast Fourier Transforms could take 30 minutes. Adding the hardware math coprocessor would bring it down to tens of seconds.

Those were the days, weren't they? But most more recent desktop-computer CPU chips have had floating-point processors build into them. The main CPU's without FPU's these days are many embedded ones.

What is the smallest consecutive 4 4th powers that add up to a 4th power?

Smallest consecutive 5 5th powers that add up to a 5th?

....

What is the general formula for the gap between their first (lowest) powers?

For 2, the first one is 3: 3^2 + 4^2 = 5^2. Those up to 1 million are {3, 20, 119, 696, 4059, 23660, 137903, 803760}.

For 3, the only one less than 1 million is 3: 3^3 + 4^3 + 5^3 = 6^3.

For 4, 5, and 6, I could find none that are less than 1 million.

For 2 and 3, the first ones are the only all-consecutive ones.