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26 is unique between a square (5^2) and a cube (3^3) 
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#1
Oct2910, 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? 


#2
Oct3010, 01:18 AM

PF Gold
P: 1,059

I think this is difficult problem. However, there is a trivial solution (1)^22=(1)^3.



#3
Oct3010, 11:42 AM

P: 258




#4
Oct3010, 02:51 PM

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?



#5
Oct3010, 04:24 PM

P: 258




#6
Oct3010, 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. 


#7
Oct3010, 10:39 PM

PF Gold
P: 1,059




#8
Oct3110, 09:19 AM

P: 39




#9
Oct3110, 08:08 PM

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. 


#10
Oct3110, 08:33 PM

P: 39

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. 


#11
Oct3110, 08:55 PM

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.



#12
Nov210, 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. 


#13
Nov210, 09:44 PM

P: 258

Is there anything not clear, or perhaps wrong? I apreciate some feedback



#14
Nov410, 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! 


#15
Nov410, 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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