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let $a,b\in \mathbb Z$, and such $$2a^2-1=b^{2013}$$

find all value $a,b$

I think $(a,b)=(0,-1),(1,1),(-1,1)$ is solution, and have other solution. Thank you everyone

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2  
As $3\mid2013$ I would try to prove non-existence of other solutions of $2a^2-1=c^3$. An elliptic curve, but probably has some trick method. –  Jyrki Lahtonen Jul 6 '13 at 13:59
    
Thank you all same, @JyrkiLahtonen –  math110 Jul 6 '13 at 14:02
    
According to me, we can prove that $\frac{b^{2013}}2 + \frac12$ is a perfect square only for the solutions you gave above then I think we might get something. –  Rohinb97 Jul 6 '13 at 14:14
    
this problem How to prove $$\dfrac{b^{2013}}{2}+\dfrac{1}{2}$$ is a perfect. –  math110 Jul 6 '13 at 14:16
1  
If it is from an ongoing competition (as seems possible given the appearance of 2013), you should tell us (and wait until deadline has passed). Otherwise it may get ugly, allegations of cheating and such. –  Jyrki Lahtonen Jul 6 '13 at 14:57

1 Answer 1

Following Jyrki's suggestion, set $c=b^{671}$ and the equation becomes $$2a^2=c^3+1$$ Multiply both sides by 8 to get $$(4a)^2=(2c)^3+8$$ This is a Mordell curve, which have finitely many solutions. In this case the constant is $8$, and here we can find the only integral solutions $$(4a,2c)\in \{(0,-2),(3,1),(-3,1), (4,2), (-4,2), (312,46), (-312,46)\}$$ This gives $$(a,c)\in \{(0,-1),(1,1),(-1,1),(78,23),(-78,23)\}$$

However only the first three give $c$ that is a 671st power. This confirms that there are three solutions in the integers.

Followup: We may replace $2013$ with $3(2k+1)$ for any integral $k\ge 2$ and get the same result.

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Hello, @vadim123,I can't Open the Mordell curve,can you send to me? –  math110 Jul 6 '13 at 16:24
    
tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL+ it's a text file –  vadim123 Jul 6 '13 at 16:29
    
Thank you @vadim123, where are this proof? with this $$c^3+1=2a^2$$ has only $$(a,c)\in\{(0,-1),(1,1),(-1,1),(78,23),(-78,23)\}$$? –  math110 Jul 6 '13 at 16:35
    
Mordell curves are of the form $y^2=x^3+n$, and there are only finitely many integral solutions for any fixed $n$. For more details, see the wikipedia link I posted. The second link I posted (see tnt.math.se.tmu.ac.jp/simath/MORDELL) gives the specific solutions for all $n$ with $|n|<10000$. They were computed using computer software. –  vadim123 Jul 6 '13 at 16:48
    
Thank you, I hope see this equation $x^3+1=2y^3$ integer solution methods? But I can't find it,can you link ? Thank you very much –  math110 Jul 6 '13 at 16:57

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