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26 is unique between a square (5^2) and a cube (3^3)

by al-mahed
Tags: cube, square, unique
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al-mahed
#1
Oct29-10, 10:58 PM
P: 258
i.e, [tex]a^{2} \pm 2 = b^{3}[/tex] has (25,27) as the only solution.

Now, can you prove it?
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robert Ihnot
#2
Oct30-10, 01:18 AM
PF Gold
P: 1,059
I think this is difficult problem. However, there is a trivial solution (1)^2-2=(-1)^3.
al-mahed
#3
Oct30-10, 11:42 AM
P: 258
Quote Quote by robert Ihnot View Post
I think this is difficult problem. However, there is a trivial solution (1)^2-2=(-1)^3.
Hi Robert, I'm sorry I didn't clarify that a, b must be positive integers.

Petek
#4
Oct30-10, 02:51 PM
Petek's Avatar
P: 361
26 is unique between a square (5^2) and a cube (3^3)

Are you asking for help in proving the statement or challenging us to give a proof?
al-mahed
#5
Oct30-10, 04:24 PM
P: 258
Quote Quote by Petek View Post
Are you asking for help in proving the statement or challenging us to give a proof?
second option :)
hochs
#6
Oct30-10, 04:28 PM
P: 39
It's easy if you use Z[sqrt(-2)] and Z[sqrt(2)] (they're both UFD).

For example look at a^2 + 2 = b^3. This factors as (a + sqrt(-2))(a - sqrt(-2)) = b^3, and the units of Z[sqrt(-2)] are just +/-1, so

a + sqrt(-2) = (c + d*sqrt(-2))^3 for some c,d in Z.

So

a = c^3 - 6cd^2 = c*(c^2 - 6d^2) and
1 = d*(3c^2 - 2d^2).

So d = +/-1 and c = +/- 1, whence you get a = +/-5, and so b = 2.

Same situation for a^2 - 2 = b^3.
robert Ihnot
#7
Oct30-10, 10:39 PM
PF Gold
P: 1,059
Quote Quote by hochs View Post
It's easy if you use Z[sqrt(-2)] and Z[sqrt(2)] (they're both UFD).

For example look at a^2 + 2 = b^3. This factors as (a + sqrt(-2))(a - sqrt(-2)) = b^3, and the units of Z[sqrt(-2)] are just +/-1, so

a + sqrt(-2) = (c + d*sqrt(-2))^3 for some c,d in Z.

So

a = c^3 - 6cd^2 = c*(c^2 - 6d^2) and
1 = d*(3c^2 - 2d^2).

So d = +/-1 and c = +/- 1, whence you get a = +/-5, and so b = 2.

Same situation for a^2 - 2 = b^3.
Seem O.K. to me, except b=3. In the second case we have a=+/-1 b=-1. However, Dr. Math makes it more complicated: http://mathforum.org/library/drmath/view/51569.html
hochs
#8
Oct31-10, 09:19 AM
P: 39
Quote Quote by robert Ihnot View Post
Seem O.K. to me, except b=3. In the second case we have a=+/-1 b=-1. However, Dr. Math makes it more complicated: http://mathforum.org/library/drmath/view/51569.html
Yea, I don't know why the post in that link is using both a + sqrt(-2) = w^3 and a - sqrt(-2) = t^3 (conjugate). You only need the first, you get just as much information just more quickly.
Petek
#9
Oct31-10, 08:08 PM
Petek's Avatar
P: 361
To go from

[tex](a + \sqrt{-2})(a - \sqrt{-2}) = b^3[/tex]

to

[tex]a + \sqrt{-2} = (c + d\sqrt{-2})^3[/tex]

don't we need that [itex]a + \sqrt{-2}[/itex] and [itex]a - \sqrt{-2}[/itex] are relatively prime? In fact, as Doctor Rob points out on the Math Forum, GCD([itex]a + \sqrt{-2}, a - \sqrt{-2}[/itex]) divides [itex]2\sqrt{-2}[/itex]. I think that's why he has to consider several cases.
hochs
#10
Oct31-10, 08:33 PM
P: 39
Quote Quote by Petek View Post
To go from

[tex](a + \sqrt{-2})(a - \sqrt{-2}) = b^3[/tex]

to

[tex]a + \sqrt{-2} = (c + d\sqrt{-2})^3[/tex]

don't we need that [itex]a + \sqrt{-2}[/itex] and [itex]a - \sqrt{-2}[/itex] are relatively prime? In fact, as Doctor Rob points out on the Math Forum, GCD([itex]a + \sqrt{-2}, a - \sqrt{-2}[/itex]) divides [itex]2\sqrt{-2}[/itex]. I think that's why he has to consider several cases.
Yes, that's right. So what I wrote above is not entirely correct, but one gets the idea - just factor and use UFD-ness of Z[sqrt(-2)].

More importantly, does anyone know how to use tex here without having to type itex and \itex everytime I want it? for example i prefer to $ $, just much easier.
Petek
#11
Oct31-10, 08:55 PM
Petek's Avatar
P: 361
The pdf file linked at the beginning of this thread strongly suggests that itex, /itex and its variants are the only format that works here.
al-mahed
#12
Nov2-10, 03:48 PM
P: 258
Hi folks, I followed a different direction.

first notice that a and b must have the same parity, so first case:

1) a, b = even

[tex](2k)^{2}\pm 2 = (2n)^{3}[/tex]
[tex]4k^{2}\pm 2 = 8n^{3}[/tex]
so,
[tex]2k^{2}\pm 1= 4n^{3}[/tex]

absurd, since even number cannot be equal to odd number, so second case:

2) a, b = odd

[tex](2k+1)^{2}\pm 2 = (2n+1)^{3}[/tex]

2.1) + 2

[tex]4k(k+1)+3 = 8n^{3}+12n^{2}+6n+1[/tex]

wich means

[tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4[/tex]
[tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k[/tex]
and
[tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (K+1)[/tex]

so

[tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4[/tex]
[tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod k[/tex]
and
[tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod (K+1)[/tex]

given [tex]f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .... + a_{i}x^{m_{i}}[/tex]
and [tex]g(x) = b_{0} + b_{1}x + b_{2}x^{2} + b_{3}x^{3} + ....+ b_{j}x^{n_{j}}[/tex]

we know that

[tex]f(x)\equiv\ g(x)\mod W[/tex] [tex]\Leftrightarrow[/tex]
[tex] a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W[/tex]

lets make

[tex]f(x) = 4n^{3}+6n^{2}+3n[/tex]

and

[tex]g(x) = 1[/tex]

finally (skip obvious manipulations)

[tex]12\equiv\ 0 \mod 4[/tex]
[tex]12\equiv\ 0 \mod k[/tex]
and
[tex]12\equiv\ 0 \mod (K+1)[/tex]

k must be = 2

2.2) - 2

(...) (after the same steps)

finally

[tex]14\equiv\ 0 \mod 4[/tex]

what is untrue

Any remarks would be apreciated.
al-mahed
#13
Nov2-10, 09:44 PM
P: 258
Is there anything not clear, or perhaps wrong? I apreciate some feedback
al-mahed
#14
Nov4-10, 01:22 AM
P: 258
Thinking again, I suspect I misunderstood the theorem found here:

http://planetmath.org/encyclopedia/P...ongruence.html

and jumped to a conclusion prematurely.

[tex]f(x)\equiv\ g(x)\mod W[/tex]
[tex] a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W[/tex]

Seems that the conclusion above is wrong.

Anyway, I'd preciate if anyone could make the theorem more clear to me, and make any comments about what I've posted :)

thank you!
al-mahed
#15
Nov4-10, 01:23 AM
P: 258
[tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4[/tex]
[tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k[/tex]
and
[tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (k+1)[/tex]

so

[tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4[/tex]
[tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod k[/tex]
and
[tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod (k+1)[/tex]


that's also wrong, but do not detroy the following argument because

[tex]24\equiv\ 0 \mod 4[/tex]
[tex]24\equiv\ 0 \mod k[/tex]
and
[tex]24\equiv\ 0 \mod (k+1)[/tex]

still would imply that k must be = 2


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