Hey bobbym;

I agree, -1 is not a square.

But... it is a cube.  and 1 is a square.  So this leaves 0 between a square and a cube.

Ok.. got it... you were simply pointing out that you could not believe someone did not realize that.

This might be something intersting to prove...

For all natural numbers, there exists an infinite amount of n and m's that satisfy this condition: n^2 = m^3.

a proof by induction i think would work here...

If you make the function n = (m^3)^(1/2), the solutions are

(1, 1)
(4, 8)
(9, 27)
(16, 64)
...
The independent values are all the perfect squares, the dependent values are all the perfect cubes.

aha... very interesting!  Expansion of the mind again! 

Could an induction proof flow like this somewhat? (I haven't done higher level math in a while and I am trying to remember what I did in college.)

We are trying to prove that:

s^2 = c^3 or

s = (c^3)^(1/2).

The solution sets (c, s) are of the form (n^2, n^3) for n = 1, 2,....

Indeed, when n = 1,

we have the solution set (1, 1) and 1 = 1^2 = 1^3 = 1.

In general, substituting c = n^2 and s = n^3 into s = (c^3)^(1/2)

gives us n^3 = ((n^2)^3)^(1/2) = n^3.

Does n + 1 work?

(n + 1)^3 = (((n + 1)^2)^3)^(1/2) = (n + 1)^3

Indeed it does, so there are an infinite amount of s and c that satisfy the condition s^2 = c^3.

haha now I see it...

Basically, I am trying to use a grenade launcher to kill an ant, when all I really would need to do is step on the thing!

Thanks for checking into it for me!

Zhylliolom wrote:

Theorem:

There are infinitely many primes.

Proof:

Define a topology on the set of integers by using the arithmetic progressions (from -∞ to ∞) as a basis.  It is easy to verify that this yields a topological space.  For each prime p let A_p consist of all multiples of p.  A_p is closed since its complement is the union of all the other arithmetic progressions with difference p.  Now let A be the union of the progressions A_p.  If the number of primes is finite, then A is a finite union of closed sets, hence closed.  But all integers except -1 and 1 are multiples of some prime, so the complement of A is {-1, 1} which is obviously not open.  This shows A is not a finite union and there are infinitely many primes.

Here is a much more elementary proof that there are infinitely many primes:

Let

be prime, and

. By the Fundamental Theorem of Arithmetic M has a prime factor which clearly must be greater than

.

Last edited by DrSteve (2011-01-14 01:45:59)

Interesting proofs involving cubes


1. The only integer solution to (x+y+z)^3=xyz is 0,0,0

If, of x,y,z, all are ==1(mod3) or all are ==2(mod3) then x+y+z is a multiple of 3 but xyz cannot be a multiple of 3 if none if its factors is. If all are multiples of 3 then, except in the case where x,y,z=0,0,0, dividing x,y and z by 3 will also yield a solution, so a non-trivial solution with all multiples of 3 only exists if there is a solution with at least one not a multiple of 3.

If, of x,y,z, two are ==1(mod3) and one is ==2(mod3), (x+y+z)^3== (1+1+2)^3==1(mod3) but xyz==2(mod3). If one is ==1(mod3) and 2 are ==2(mod3) then (x+y+z)^3== (1+2+2)^3==2(mod3) but xyz==1(mod3).

2.

since 1=1+1/2 and

and (n+2)(n+1)/2=n+1+n(n+1)/2 therefore all successive successors of 1 satisfy this sum (tounge twister ).

since 1^3= ((1+1)/2)^2 and

and

.

Last edited by namealreadychosen (2011-08-01 18:18:49)