Solve in positive integers the following equation: \[{1\over n^2}-{3\over 2n^3}={1\over m^2}\] Proposed by A. Golovanov
Problem
Source: Tuymaada 2012, Problem 5/6, Day 2, Seniors/Juniors
Tags: number theory unsolved, number theory
anonymouslonely
18.07.2012 13:08
$ m^{2}=\frac{2n^{3}}{2n-3} $ and $ (2n-3,2n^{3})=3 $ so $ 2n-3=1,3 $. so the only solution is $ n=2 $.
mavropnevma
18.07.2012 13:30
That claim is in fact false. For $n=6$ we have $\gcd(2n-3,2n^3) = 9$. If we also allow negative integers, there is another case $n=-12$, when $\gcd(2n-3,2n^3) = 27$. All possible cases are $2n-3\in \{-27,-9,-3,-1,1,3,9,27\}$, but none leads to a solution, except $n=2$, $|m|=4$ (even if negative integers are also allowed).
anonymouslonely
18.07.2012 13:52
sorry...what a terrible mistake in fact $ (2n-3,2n^{3}) $ divides $ 27 $. so $ 2n-3=1,3,9,27 $. then $ n=2,3,6,15 $. but $ n $ must be even and $ n=6 $ is not a solution. so the only one is $ n=2 $.
Ta_ka
18.07.2012 14:45
i think fermats infinite descent on the degree of 2 in $m^2$ and $2n^3$ kills it instantly
mavropnevma
18.07.2012 14:58
No, it does not. Let $m=2^a x$ and $n = 2^b y$, with $x$ and $y$ odd. Then we need $2^{2a}x^2(2^{b+1} y - 3) = 2^{3b+1} y^3$, whence $2a = 3b+1$. This forces $b=2c-1$ to be odd (for some $c\geq 1$), thus $a=3c-1$, and the powers of $2$ disappear from the equation, which now writes $x^2(4^{c} y - 3) = y^3$. Gone is Fermat's infinite descent ... In fact, this is the start of an alternative solution; we need now $4^{c} y - 3 \mid y^3$, and via similar $\gcd$ considerations we again reach $n=2$, $m=|4|$ as sole integer solution(s).
mavropnevma
21.07.2012 15:35
The equation writes $m^2(2n-3) = 2n^3$. Let $m = 2^a x$, $n=2^b y$, with odd $x,y$. We then need have $2a = 3b+1$, thus $a = 3t-1$, $b=2t-1$, for some integer $t\geq 1$. The equation now writes $x^2(2^{2t} y - 3) = y^3$. But $\gcd(2^{2t} y - 3, y) \mid 3$.
$\bullet$ If $\gcd(2^{2t} y - 3, y) = 3$ we then need $y=3z$, and so $x^2(2^{2t} z - 1) = 3^2z^3$. But $\gcd(2^{2t} z - 1, z) = 1$, and so we need $2^{2t} z - 1 \mid 3^2$. There are only two potential cases
$\quad \quad \quad \bullet$ $2^{2t} z - 1 = 3$, for $t=1$ and $z=1$, leading to $x^2 = 3$, impossible;
$\bullet$ $2^{2t} z - 1 = -9$, for $t=1$ and $z=-2$, but $z$ is odd, so impossible.
$\quad \quad \quad \bullet$ If $\gcd(2^{2t} y - 3, y) = 1$ we then need $2^{2t} y - 3 = \pm 1$, hence $2^{2t} y = 4$. This leads to $t=1$, $y=1$, so $x= \pm 1$, therefore $m = \pm 4$ and $n=2$.
Alternative Solution.
The equation writes $m^2 = \dfrac {2n^3} {2n-3}$, and again $(2m)^2 = \dfrac {8n^3} {2n-3} = (2n-3)^2 + 9(2n-3) + 27 + \dfrac {27} {2n-3}$. Thus one only has to check $2n-3\in \{-27,-9,-3,-1,1,3,9,27\}$ (we can quickly eliminate half of them, since clearly $n$ must be even), and the only one that works is $2n-3=1$, when $n=2$ and $m=\pm 4$.
Lawasu
24.07.2012 15:37
mikolez wrote: Solve in positive integers the following equation: \[{1\over n^2}-{3\over 2n^3}={1\over m^2}\] The equation becomes $(2n-3)m^2=2n^3$, so $2\mid m^2$, hence $m=2z$ where $z$ is a positive integer. Now we get $4(2n-3)z^2=2n^3$, so $2\mid n^3$, hence $n=2x$ where $x$ is a positive integer. Next, $(4x-3)z^2=4x^3$, so $2\mid z$, hence $z=2y$ where $y$ is a positive integer. The equation becomes $(4x-3)y^2=x^3$. Take $d=gcd(x,y)$ and let $a,b$ be positive integers such that $x=da$ and $y=db$. Then $gcd(a,b)=1$. We have $(4da-3)b^2=da^3$. It follows that $a\mid 4da-3\Rightarrow a\mid 3$. If $a=1$, then $(4d-3)b^2=d$, If $d>1$, then $(4d-3)b^2\ge 4d-3>d$. Therefore $d=1$ and $b=1$. We get . If $a=3$, then $(4d-1)b^2=9d$. If $b\ge 2$, then $(4d-1)b^2\ge 3db^2>9d$. Therefore $b=1$, but in this case $d$ can't be an integer. Finally, we obtain the solution $(m,n)=(4,2)$. It isn't necessary to solve this with Fermat Infinite Descending.