u,v,w are Z+

Perhaps someone brainy can recognize the above equation. I've tried to find a general pattern to ALL the solutions but have failed so far. It's not an easy pattern.

Can anyone give me an infinite number of solutions? If not, then whats the largest solution you can find? (largest w)

I don't know the pattern of the roots of the equation but if you write the equation like this:

u(u^2 - 1) + v(v^2 - 1) = w(w^2 - 1)

you get

(u - 1)u(u + 1) + (v - 1)v(v + 1) = (w - 1)w(w + 1)

(x-1)x(x+1) is the product of 3 following numbers, so you can write that as

[(u+1)!]/[(u-2)!] + [(v+1)!]/[(v-2)!] = [(w+1)!]/[(w-2)!]

and if you multiply everything by 1/(3!)
you'll get

[(u+1)!]/[3!*(u-2)!] + [(v+1)!]/[3!*(v-2)!] = [(w+1)!]/[3!*(w-2)!]

and this equals

_3______3______3
C____+ C____= C
_u+1____v+1____w+1

(binomial coefficients)
(Ignore the underlines. I used them coz the spaces get trimmed)

so u^3-u+v^3-v=w^3-w becomes the above equation.

So the equation has an infinite number of solutions.
I guess you've probably figured that out...
Hope it helps...

How do you know it has infinite number of solutions?

Gerbile,
looks nice, well done, but this:
"if you write the equation like this:

u(u^2 - 1) + v(v^2 - 1) = w(w^2 - 1)

you get

(u - 1)u(u + 1) + (v - 1)v(v + 1) = (w - 1)w(w + 1)

(x-1)x(x+1) is the product of 3 following numbers"
seems enough to show there are an infinite number of solutions, becuase it is p + q = r, with p=(u - 1)u(u + 1), etc.

Merlijn,

it may be p+q=r, but with a stipulation that each of p, q and r is factorable into three consecutive integers...? How do you guarantee that r can be arbitrarily large? This sort of smells like finding the largest prime, which is an unsolved problem.

the mmoment I posted I realised that it was a bit too thin! Well, I'll just bame this being pre-coffee time of the day. :)

but with regard to the prime thing:
y = x!+1
1< z <= x

y/z = (x!+1)/z = x!/z + 1/z = [(z-1)!z(x!/z!)]/z + 1/z = (z-1)! (x!/z!) + 1/z.

(z-1)!(x!/z!) is always element of N
this means that y = x!+1 must be a prime, or can only be devided by numbers larger than x, making those numbers primes lager than x.
x=5; 5!+1 = 121 is not a prime , but can devided by 11 (11>5 and is a prime)
or x=3; 3!+1=7, 7 is a prime.

Since x can be any number, there is no largest prime.

Ooooh, that was beautiful. Thanks, Merlijn! (I must've been thinking of some other unsolved problem....) :confused:

"Ooooh, that was beautiful. "
thanks.... I must admit, I cried when I first wrote down the proof.
I once read (in Goedel, Escher, Bach) that Euclides had proven there are an infinite amount of primes. All Hofstadter said that it came down to the x!+1. If you have that hint, it's quite easy to make up the rest, still the sheer elegant beauty of the proof overwhelemed me for a second.

Merlijn, thanks for the remark. I didn't notice that.

I've noticed something that might help in finding the pattern of the answers.
The product of three consecutive integers (thanks for widening my vocabulary, Overdoze :)) can be also represented using a sum of factorials.
for example:
u=4=>3*4*5=60=4!*2+3!*2
u=5=>4*5*6=210=5!+3*4!+3*3!
...

It's pretty obvious considering the relation of the equation to the binomial coefficients.

Just in case you didn't notice.

Originally posted by BloodSuckingGerbile

The product of three consecutive integers (thanks for widening my vocabulary, Overdoze :)) can be also represented using a sum of factorials.
for example:
u=4=>3*4*5=60=4!*2+3!*2
u=5=>4*5*6=210=5!+3*4!+3*3!
...


Glad to be of service. :)

But...what the heck are you talking about??

5! alone is 2*3*4*5 = 4*5*6, which is by the way 120 not 210

:bugeye: :eek: :confused:

:eek:
sorry, I meant
u=6=>5*6*7=210=5!+3*4!+3*3!

Ok, sorry for being so dense. :o

And I apologize again, but after hours of contemplation I don't see what you're describing. i.e. I don't see a pattern of factorial sums for 3 consecutive integer products. Could you please post a general formula you're using, or even better yet a derivation?

As an aside:

x^n + y^n = z^n

is Fermat's last theorem, which says there are no integer solutions for n > 2. Though I'm not sure how that bears, if at all, on the particular form presented in this thread...

:confused:

I didn't find a pattern either, I just noticed it. I didn't find a formula for it or anything, however I'll try to derive it.
I hope the derivation is not as long as for Fermat's theorem...
by the way, why did you post it (Fermat's theorem)?:confused:

I didn't find the pattern of the factorials, however I did notice that the integers formed by the product of three consecutive numbers

(u-1)u(u+1)

can be written as

3!*C_n

where n=u-1 (u>1)

It's obvious, because the product is equal to

(u+1)!/(u-2)!

and if you divide that by 3!, it will form a binomial coefficient, which is always N.

So C_nn is a series where

C_n+1=C_n+b_n

and b_n+1=b_n+n+2

b₁=3.
c₁=1.

I doubt that this would help in finding the pattern of all the solutions, but.. who knows?

The reason for my mentioning Fermat, is that it is very similar in form to the problem posed here. Fermat says that

u^3 + v^3 = w^3

has no Z+ solution sets. Rewriting the above equation, you have

u*u*u + v*v*v = w*w*w

In the problem posed in this thread, you have

(u-1)*u*(u+1) + (v-1)*v*(v+1) = (w-1)*w*(w+1)

The forms are eerily similar. Which made me wonder if the equation has any Z+ solutions at all, let alone an infinite number of them. I don't know what I was thinking, since obviously there *are* solutions. e.g. w=5, u=v=4

[edit: removed some raving lunacy I've included previously. Must've been something I ate...]

Itchy, I think Overdoze was right and we were all wrong. The series I was talking about is really weird and I doubt that the sum of any C_u
with any other C_v will be equal C_w.
I'll try to prove that and then I'll tell you for sure.

Not everything is as obvious as it seems...

.......
Today I was really bored so I derived the formula for C_n.

C_n=(n³+3n²-16n+12)/6 +3n-2.

C₁=1
C₂=4
C₃=10
C₄=20
C₅=35
...

And then I wrote
C_n+C_n+k
in order to prove that a sum of two terms of the series is not equal another term in the same series.
That is eventually equal
6(2C_n+C_k)+3nk(2+n+k).
But that doesn't prove it ...or does it? :confused:
Work with me people!!