An Interesting question that yet lies unanswered (1 Viewer)

[Lucas Shine]

We were given an interesting question the other day


a^3+b^3+c^3+d^3=100^100

Find a, b, c and d.


I have wrestled a bit with it and have only managed to come up with something really general like d=-(a^3+b^3+c^3-10^200)^1/3.

This may be a famous question and may already have been answered somewhere, but I was wondering if you could point me in the right direction or link to a site/thread that has the answer.

I'm at 4 unit level in year 12 so that gives you an idea of where I'm at in terms of knowledge base, although I don't mind you posting university stuff as I don't mind being challenged a bit.


Thanx in advance
lucas

 

[tommykins]

you can't explicitly find a,b,c and d (i.e assign an actual value to it) seeing as there is an infinite amount of solutions (especially in ^3 where negatives come into play) which is my assumption just looking at it.

your solution is realistically the best bet.

 

[Lucas Shine]

I forgot to mention, they must be positive integers (as far as I remember). Is it posible to find one combination of numbers? If so, could you show me how?

 

[Lucas Shine]

hmm... does it work if a=10^66, b=2(10^66), c=3(10^66) and d=4(10^66)

A friend is telling me that I'm close but wrong. Can you verify either his or my statement?


Appreciate it

 

[tommykins]

I personally can't see an optimal way of doing it without trial and error.

Noting 100^100 = (10^2)^100 = 10^200

(10^66)^3 = 10^198 so it is possible that you're close, altough with your abcd values you'd get 10^198(1+8+27+64) = 100*10^198 = (10^2).(10^198) = 10^200 = 100^100 as above..

maybe an error in there, but you should be right.

 

[Iruka]

100^100=100 x 100^99 = 100 x (100^33)^3

100 = 1+8+27+64

i think you can work it out from there.

 

[Affinity]

what about uniqueness/knowing that all solutions are found

 

[Iruka]

Well, I never claimed that it was the only solution.

There are lots of theorems about writing numbers as the sum of squares, but I don't know anything about sums of cubes.

Actually, I think the solution illustrates the fact that
(1+2+3+...+n)^2=1^3+2^3+...+n^3

So any triangular number squared can be written as the sum of n cubes.

 

[pman]

there are infinite solutions, it you introduce parameters for 3, you can find the last in relation to those parameters but thats about it

 

[Affinity]

it's finite.. all of them have to be positive integers

 

[tommykins]

it's finite.. all of them have to be positive integers

If they could be negative however, without rigorous proof - is it safe to say that there possibly exists infinite solutions?

 

[Archman]

According to this site: Sum of four cubes

n = 100 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000
000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000
000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000
000000 (201 digits)

n = a^3 + b^3 + c^3 + d^3
a = 5 555555 555555 555555 555555 555555 555555 555555 555555 555555 555555
555555 555555 555555 555555 555555 555555 555555 555555 555555 555555 555555
555555 555555 555555 555555 555555 555555 555555 555555 555555 555555 555555
555561 (199 digits)
b = -5 555555 555555 555555 555555 555555 555555 555555 555555 555555 555555
555555 555555 555555 555555 555555 555555 555555 555555 555555 555555 555555
555555 555555 555555 555555 555555 555555 555555 555555 555555 555555 555555
555570 (199 digits)
c = -16 666666 666666 666666 666666 666666 666666 666666 666666 666666 666666
666666 666666 666666 666666 666666 666666 666666 666666 666666 666666 666666
666666 666666 666666 666666 666666 666666 666666 666666 666666 666666 666666
666697 (200 digits)
d = 16 666666 666666 666666 666666 666666 666666 666666 666666 666666 666666
666666 666666 666666 666666 666666 666666 666666 666666 666666 666666 666666
666666 666666 666666 666666 666666 666666 666666 666666 666666 666666 666666
666698 (200 digits)


Read Cubic Number -- from Wolfram MathWorld for more info about expressing numbers as sum of four cubes. I suspect that the question you're given was not an exercise. Probably to provoke thoughts on whether it is always possible to express numbers in terms of sum of cubes etc.

Also read up on expressing numbers on sum of four squares too! Interesting stuff.