Fermat Last Theorem look-alike

[Hyperreality]

I encountered this problem when I was reading Dr Riemann's Zero, the problem is:

Find the integers a, b and c such that a^3 + b^3 = 22c^3.

First I did some research from the public library and obtained some lecture notes on Fermat's Last Theorem, but I made little progress, partly because I don't understand most part of the book.

The only thing I know about this equation is that they are all relative primes, and a^3 and b^3 must be odd numbers.

Can somone please show me how this is done?

[Hurkyl]

A lot of times, when I've seen diophantine equations solved directly, one plugs what they know back into the equation and tries to elicit new information.

In this case, you know a and b must be odd, so write a=2d+1 and b=2e+1, expand, simplify, and see if you can discover something else.



(p.s. are you searching for solutions or trying to prove impossibility?)

[QuantumNet]

sorry, I tried to solve it. I'll try again someday.

[MathematicalPhysicist]

Originally posted by QuantumNet
7a^3 + 7b^3 + 7ab^2 + 7ba^2 = ab^2 + ba^2


it should be: 7a^3+7b^3+21ab^2+21ba^2=ab^2+ba^2

[QuantumNet]

Originally posted by loop quantum gravity
it should be: 7a^3+7b^3+21ab^2+21ba^2=ab^2+ba^2

I corrected that in a few minutes, you're a fast commenter.

[robert Ihnot]

Need ideas! Any ideas?

Cubic Problem X^3 + Y^3 = 22Z^3 a problem.

Considering the method of Euler-http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html#40-
we are able to achieve the form:

(X+Y) (X^2-XY + Y^2) = (2m) (m^2+3n^2) by setting X = m +n, Y =m-n.

This form is closed under multiplication, that is:

(s^2+3t^2)(u^2+3v^2) = (su-3tv)^2 + 3(ut + sv)^2,

We now cube the form a^2+3b^2. This results in:

[(a^2-3b^2)a-3b(2ab)]^2 + 3[(a^2-3b^2)b + a(2ab)]^3

Setting this equal to m^2 + 3n^2 and simplifying we have a cube where:

m = a(a^2-9b^2). n= 3b(a^2-b^2)

Example: Let us suppose a=1, b=2, then m=-35, n=-18. Thus:

m^2 +3n^2 =35^2 +3x18^2 = 2197= 13^3.

While a=2, b=3 leads to

154^2 + 3x45^2 = 29791 = 31^3.

Now we want to take in consideration 2m, putting it all together for a=2, b=1, we get m=-10, n=9, 2m=-20, X=-1, Y=-19.

X^3 +Y^3 = 2m(Z^3) = 1^3 + 19^3 = 20(7^3).

But we really want to find 2m = 2x11xv^3. But this is a real problem. Some results are found: 51^3 + 111^3 = 6x63^3, and 152^3 + 8^3 =20x56^3, and 67881^3 + 147741^3 = 6x(7623x11)^3.

By using a=8, b=1, we can get m==0 Mod 11, but this also satisfies a == 3 Mod 5. This approach gives us a “near miss”:

629^3+251^3 = 110(134)^3.

I am out of ideas.

[Gokul43201]

Haven't really read all the posts here but here's something that may or may not be useful.

If solutions exist, then a,b would have to be of the form 11k + n, 11m - n with n in [1,10].

This results from solving ~~a^3 + b^3 \equiv 0 ~~(mod~11)

[Muzza]

I've seen this discussed on sci.math, I don't think anyone there had a solution though...

[epte]

I ran into this problem yesterday in the introduction to The Riemann Hypothesis: The Greatest Unsolved Problem in Mathematics by Karl Sabbagh. I think I just solved it, but if I've made a mistake somewhere, I'd appreciate someone pointing it out.

The version of the problem I have is: Find positive integers a, b, and c such that a^3 + b^3 = 22c^3

I took the approach of Gokul43201, except that I centered both (11k + n) and (11m - n) around the same multiple of 11. Thus, in my notation, a = 11x - k, b = 11x + k, where k can be greater than 11. Also, b - a = 2k, which is needed to assure that a^3 + b^3 is even.

So, we have:
(11x - k)^3 + (11x + k)^3 = 22c^3

Expanding and collecting terms, we have:

121x^3 + 3xk^2 = c^3

From this, we can see that x must be a factor of c. That is, there must exist positive integer y such that xy = c. Thus, we have:

121x^3 + 3xk^2 = (xy)^3
3xk^2 = x^3 * y^3 - 121x^3
3k^2 = x^2 * y^3 - 121x^2
3k^2 = x^2(y^3 - 121)

From this, we can see that x^2 must be a factor of 3k^2, and x must be a factor of k. There must exist positive integer m such that mx = k. Therefore, we can reduce to:

3m^2 = y^3 - 121

We've eliminated a variable (is this problematic? I was hoping not).

Now, the right side must be divisible by 3. Since 121 (mod 3) = 1, then y^3 (mod 3) = 1, and there must exist positive integer z such that (3z + 1) = y. Thus,

3m^2 = (3z + 1)^3 - 121
3m^2 = 27z^3 + 27z^2 + 9z - 120
m^2 = 9z^3 + 9z^2 + 3z - 40
m^2 + 40 = 3z(3z^2 + 3z + 1)

From this, we see that m^2 + 40 must be divisible by 3. Since 40 (mod 3) = 1, that means m^2 (mod 3) = 2. But this is impossible. Consider, if x (mod 3) = 1, then x^2 (mod 3) = 1. If x (mod 3) = 2, then x^2 (mod 3) = 1. Therefore, there exist no positive integers a, b, and c such that a^3 + b^3 = 22c^3

This is the first time I've attempted a proof of this sort, so I've probably made a mistake somewhere, and am looking forward to being corrected.

Thanks,
Erich Enke

[Hurkyl]

I took the approach of Gokul43201, except that I centered both (11k + n) and (11m - n) around the same multiple of 11. Thus, in my notation, a = 11x - k, b = 11x + k, where k can be greater than 11. Also, b - a = 2k, which is needed to assure that a^3 + b^3 is even.
In case you missed it, you needed to know that b-a is even in order to center them both around the same multiple of 11. e.g. if a=1 and b=10, they are of the form a=11k-n and b=11m-n, but cannot be written in the form you used.

[epte]

In case you missed it, you needed to know that b-a is even in order to center them both around the same multiple of 11. e.g. if a=1 and b=10, they are of the form a=11k-n and b=11m-n, but cannot be written in the form you used.

Correct. I think that was in my reasoning process, but I reported it a bit awkwardly. To put it in other words, since a^3 + b^3 must be even, the difference between a and b must be even. a and b therefore cannot be (11x + 1) and (11x + 10) respectively, since their difference will always be odd.

[epte]

Hm.. but I do agree with you -- that's a bit of a weak point in the proof. Am I handling all the cases when I center around the same multiple of 11? Or are there some legitimate cases I'm not considering, such that I'm only proving that there is merely no solution within a subset of possible solutions? I'll give some thought to that.

Thanks,
Erich

[epte]

Ok, how about this:

Assume a = (11d + n) and b = (11e + 11-n). Knowing the difference of a and b must be even, then:

(11d + n) - (11e + 11 - n) = 2k
11(d - e - 1) = 2(k - n)
d - e must always be odd.

Instead of using e, if we choose f such that we can use -n instead of 11-n, then we have

(11d + n) - (11f - n) = 2k
11(d - f) = 2(k-n)
d-f is always even.

If d-f is even, there always exists an integer midpoint m between d and f such that m=(d+f)/2 and d-m = (d-f)/2 and m-f = (d-f)/2

Therefore, we can always write (11d + n) - (11f - n) = 2k in terms of m and a radius r=(d-f)/2:
11(m+r) + n - 11(m-r) - n = 2k

since m+r = (d+f)/2 + (d-f)/2 = d and m-r = (d+f)/2 - (d-f)/2 = f

Therefore, every considerable pair of (mod 11) numbers can be represented as equidistant from a central multiple of 11.

[Hurkyl]

Shorter derivation: the desired multiple of 11 is the average of a and b, which is an integer iff their sum is even.