AUSTRALIAN MATHEMATICAL OLYMPIAD
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Determine all the positive integer solutions of the equation $ (x+1)^4-(x-1)^4=y^3 $.
When you are describing an equation, mention every variables, please. Otherwise it makes confusions.
When you are describing an equation, mention every variables, please. Otherwise it makes confusions.
Re: AUSTRALIAN MATHEMATICAL OLYMPIAD
Here's my solution::
$((x+1)^2)^2)-((x-1)^2)^2)=y^3$
$\Rightarrow (2x^2+2)(4x)=y^3$
$\Rightarrow x(x^2+1)=k^3$ [where, $y=2k$]
If $x$ only has integer solutions, $k$ only has integer solutions.
Now, trial and error implies that there's no positive integer solutions for $(x,y)$ .
Let me know if there's any bug in the solution. (I somewhat have a strong feeling that there's bug in it... )
$((x+1)^2)^2)-((x-1)^2)^2)=y^3$
$\Rightarrow (2x^2+2)(4x)=y^3$
$\Rightarrow x(x^2+1)=k^3$ [where, $y=2k$]
If $x$ only has integer solutions, $k$ only has integer solutions.
Now, trial and error implies that there's no positive integer solutions for $(x,y)$ .
Let me know if there's any bug in the solution. (I somewhat have a strong feeling that there's bug in it... )
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Re: AUSTRALIAN MATHEMATICAL OLYMPIAD
I want to add from here.:Labib wrote:Here's my solution::
$((x+1)^2)^2)-((x-1)^2)^2)=y^3$
$\Rightarrow (2x^2+2)(4x)=y^3$
$\Rightarrow x(x^2+1)=k^3$ [where, $y=2k$]
$g.c.d.(x,x^2+1)=1$ So, set $x=m^3,x^2+1=n^3$ such that $mn=k$ and $(m,n)=1$.
Now, it implies, $1=(n-m^2)(n^2+nm^2+m^4)\geq 3$ as $n,m\geq1$. That's why we have no solution.
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: AUSTRALIAN MATHEMATICAL OLYMPIAD
But in rational positive number, infinite solution exist.
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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Re: AUSTRALIAN MATHEMATICAL OLYMPIAD
$ (x+1)^4-(x-4)^4$=$ ((x+1)^{2})^{2}$ - $((x-1)^{2})^{2} = ((x+1)^{2}+(x-1)^{2} )( (x+1)^{2}-(x-1)^{2} )=2(x^2+1)*(4.x.1)=8x^3+8x.$.
Now,$ (2x)^3<8x^3+8x<(2x+1)^3.$
or, $(2x)^3< (x+1)^4-(x-4)^4<(2x+1)^3.$
or,$(2x)^3<y^3<(2x+1)^3.$
or,$2x<y<2x+1. $
so, hence there is no integer number between 2x & (2x+1) ; so, no solution.
Now,$ (2x)^3<8x^3+8x<(2x+1)^3.$
or, $(2x)^3< (x+1)^4-(x-4)^4<(2x+1)^3.$
or,$(2x)^3<y^3<(2x+1)^3.$
or,$2x<y<2x+1. $
so, hence there is no integer number between 2x & (2x+1) ; so, no solution.
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Re: AUSTRALIAN MATHEMATICAL OLYMPIAD
hi,friends, please ,help me. 1 of my friend today showed me that (0,0) is a solution. but, I have proved there exists no solution.So, I am wrong. What is my wrong?? I can't find out. PLZ, HELP ME for finding the wrong.
Re: AUSTRALIAN MATHEMATICAL OLYMPIAD
$x=0 \Rightarrow (2x)^3=8x^3+8x$MATHPRITOM wrote:$ (2x)^3<8x^3+8x<(2x+1)^3.$
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Nur Muhammad Shafiullah | Mahi
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Re: AUSTRALIAN MATHEMATICAL OLYMPIAD
Great.. Mahi , the matter is too simple,but, so interesting .. & I don't understand the bug.*Mahi* wrote:$x=0 \Rightarrow (2x)^3=8x^3+8x$MATHPRITOM wrote:$ (2x)^3<8x^3+8x<(2x+1)^3.$