Last week, my teacher asked us to prove this: Prove that $(2−1)^2−4$ is not a perfect cube for any integer I tried this approach:$y=2k$ and $y=2k+1$ but I couldn't prove it this way.
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Hint: Letting $z=2y-1$ you have $n^3=z^2-4=(z-2)(z+2)$. Note that $\gcd (z-2,z+2)$ must be a divisor of $4$. – lulu Nov 20 '20 at 17:40
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I don't understand why $gcd(z-2,z+2)|4$? – Emma Hoffman Nov 20 '20 at 17:43
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2Hmm... same teacher as in this question? https://math.stackexchange.com/questions/3914956/proof-not-a-perfect-square – Deepak Nov 20 '20 at 17:45
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$\gcd(z+2,z-2)=\gcd(z+2,z+2-(z-2))=\gcd(z+2,4)$ – J. W. Tanner Nov 20 '20 at 17:45
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Say $d=\gcd(a,b)$. Then $d,|,a$ and $d,|,b$ so $d$ divides, say, $b-a$. – lulu Nov 20 '20 at 17:45
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thnx I got now. but still I don't know how to use that hint. – Emma Hoffman Nov 20 '20 at 17:56
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should I replace $(−2)(+2)$ with $gcd(z-2,z+2)*lcm(z-2,z+2)$? – Emma Hoffman Nov 20 '20 at 18:08
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1ok, i will try to give the right hint... We have two numbers, $A, B\in\Bbb Z$, and their product is a perfect cube in $\Bbb Z$. Can it be that $A,B$ are not each a perfect cube? Can we exclude this from the landscape in the given situation... Feel free to answer and accept the own question! @ John Sparrow – dan_fulea Nov 20 '20 at 18:17
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My answer contained a mistake. Now I edited and it's OK – Raffaele Nov 20 '20 at 20:04
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$$(2 y - 1)^2 - 4=(2y-1+2)(2y-1-2)=(2y+1)(2y-3)$$ $(2y-3)$ and $(2y+1)$ are coprime. For euclidean algorithm
$\gcd (2y+1,2y-3)=\gcd(2y+1,4)=1$
because $(2y+1)$ is odd and is coprime with $4$
So the two numbers have no common factors. Thus or both are cubes or their product cannot be a cube.
As the difference is $4$ we see that $2y-3$ and $2y+1$ cannot be both cubes because beside the case $1^3-0^3=1$ the minimum difference between two cubes is $2^3-1^3=7$
Therefore $(2 y - 1)^2 - 4$ cannot be a cube.
Raffaele
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@PM2Ring I edited the answer the day before yesterday. $\gcd$ is 1 because the number are coprime – Raffaele Nov 21 '20 at 16:48