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#32
07-22-2005, 06:35 PM
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Re: Solved!!
x^2 + 68 = y^5 then
x^4 + 136x^2 + 68^2 = y^10 Now working modulo 11 By Little Fermat theorem, y^10 = 1 if (y,11)=1 68=2 68^2=4 136 = 4 then x^4 + 4x^2 + 4 = 1 mod 11 (x^2 + 2)^2 = 1 mod 11 therefore x^2 + 2 = 1 so x^2 = -1 mod 11 or x^2 + 2 = -1 this is x^2 = -3 = 8 mod 11 contradiction in both cases!!! Now, if y = 11k then y = 1 mod 10 so, x^2 + 68 = 1 mod 10 x^2 - 2 = 1 mod 10 x^2 = 3 mod 10 contradiction !!! Q.E.D. David 
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#33
07-22-2005, 06:39 PM
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Re: Solved!!
I am an idiot, I confess [img]/images/graemlins/smile.gif[/img].
nh.
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#38
07-22-2005, 07:26 PM
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Re: None of this nursery school stuff - a proper maths problem. 25$ reward
Here is a relevant reference. I don't know whether this contains the solution, but it reduces the problem to a finite search:
<ul type="square">MR1259344 (94k:11037) Cohn, J. H. E.(4-LNDHB) The Diophantine equation $x\sp 2+C=y\sp n$. Acta Arith. 65 (1993), no. 4, 367--381. 11D61 -------------------------------------------------------------------------------- It follows from the theory of linear forms in logarithms that for a nonzero integer $C$, the equation (1) $x^2+C=y^n$ in integers $x,y,n\geq 2$ implies that $\max$ $(|x|,|y|,n)$ is bounded by an effectively computable number depending only on $C$. The value of this number turns out to be very large. For several values of $C$, all the solutions of equation (1) have been determined. For example, if $C=11,17,40$, the only solutions of (1) are given by $x=4$ and $x=58$, $x=8$ and $x=52$, respectively. Furthermore, 46 values of $C\leq 100$ are given for which equation (1) has no solution. The paper contains an account of earlier results on equation (1).[/list]
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#40
07-22-2005, 09:18 PM
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Re: None of this nursery school stuff - a proper maths problem. 25$ reward
Hey,
Some good efforts, particularly from Sirio (& Gump). Arfinn you've got intuition I suspect, but this problem is too hard to be able to intuitively 'suspect' there are/arent solutions (unless your life is diophantine equations, then maybe). Modulo reduction and arithmetic is a useful tool in number theory, only trouble is that its limited mostly to proof by contradiction, not much good if there are actually roots! But in fact there arent any roots in this one, so if you look hard enough it might be possible to find a series of reductions to prove the non-existance of solutions. I dont know how far one has to look (potentially for ever) because I havnt proved it this way. In fact the problem is not easy, and I think in hindsight perhaps too difficult, so I'll give a very rough solution tomorrow if anyone's interested. For those still thinking, here are some clues: <font color="white"> Work in the field Z[sqrt(-17)]: (that is to say, all numbers in this field have the form 'a + b.sqrt(-17)' where a and b are in Z. Then the equation is x+2.sqrt(-17) * x-2.sqrt(-17) = y^5 Lets let A = x + 2.sqrt(-17). Now it is possible to show, by considering the factorisation of y^5, that A itself is a fifth power in this field. This is difficult. But it is possible, and indeed: A = (c + d.sqrt(-17))^5 with c and d integers. From this, you can show there are no solutions by expanding and looking for a condradiction. </font>
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