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find x

Albert

Well-known member
Jan 25, 2013
1,225
x and k are positive integers ,satisfying :

$ \text {x(x+1)=64} \text { k}^4 -48 \text{ k}^3 +16\text { k}^2-2\text{ k} $

please find x
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
x and k are positive integers ,satisfying :

$ \text {x(x+1)=64} \text { k}^4 -48 \text{ k}^3 +16\text { k}^2-2\text{ k} $

please find x
Your RHS can be factorized as $2k(4k-1)(8k^2-4k+1)$.
Trying it out for k=1, we find:
$2k(4k-1)(8k^2-4k+1) = 2 \cdot 3 \cdot 5 = 5 \cdot 6$​

So I've found $x=5$. $\qquad \blacksquare$

Found it! Phew! (Whew)
 

Albert

Well-known member
Jan 25, 2013
1,225
ILikeSerena: you got the answer (Yes)

now I think you should point out if k>1 then x will not exist

what do you think ?
 

Albert

Well-known member
Jan 25, 2013
1,225
$2k(4k-1)(8k^2-4k+1)=(8k^2-2k)(8k^2-4k+1) $

$\text{if} \,\, 8k^2-4k+1 < 8k^2-2k , \,\text{then}\, k>\dfrac {1}{2} $

let :

$ x=8k^2-4k+1 , \, x+1=8k^2-2k $

we have :

$ 8k^2-4k+2=8k^2-2k $

$\therefore k=1 , \, \, x=5 $
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I don't get it.
How do you know that x does not exist for k > 1?
 

Albert

Well-known member
Jan 25, 2013
1,225
I don't get it.
How do you know that x does not exist for k > 1?
please take note of my previous post as follows :
Albert said:
[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT]

[FONT=MathJax_Main]if[/FONT][FONT=MathJax_Main] 8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main]then [/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]1/2[/FONT]

let :

[FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]k[/FONT]

we have :

[FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]k[/FONT]

[FONT=MathJax_AMS]∴[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]5[/FONT]
$ 8k^2-4k+2=8k^2-2k -------------------(*) $

the only value for k satisfying (*) is k=1
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
please take note of my previous post as follows :


$ 8k^2-4k+2=8k^2-2k -------------------(*) $

the only value for k satisfying (*) is k=1
But how did you get $x=8k^2−4k+1$ then?
Isn't that just a try-out?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
$2k(4k-1)(8k^2-4k+1)=(8k^2-2k)(8k^2-4k+1) $

$\text{if} \,\, 8k^2-4k+1 < 8k^2-2k , \,\text{then}\, k>\dfrac {1}{2} $

let :

$ x=8k^2-4k+1 , \, x+1=8k^2-2k $

we have :

$ 8k^2-4k+2=8k^2-2k $

$\therefore k=1 , \, \, x=5 $
What this shows is that if the integer $N(k) \overset{\text{def}}{=} 64k^4 - 48k^3 + 16k^2 -2k$ is factorised as the product of the two factors $8k^2-2k$ and $8k^2-4k+1$, then the only way for those factors to differ by $1$ is if $k=1$. That still leaves open the possibility that $N(k)$ could be factorised in some different way as the product of two integers differing by $1$. So I agree with ILikeSerena that the above argument does not rule out the possibility of other solutions (though admittedly it seems unlikely that there are any).
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
So I agree with ILikeSerena that the above argument does not rule out the possibility of other solutions (though admittedly it seems unlikely that there are any).
I checked up to k=100000 with N(100000) = 6399952000159999800000.
No other solutions though.
 

Albert

Well-known member
Jan 25, 2013
1,225
as you said my RHS can be factorized as :
$2k(4k-1)(8k^2-4k+1)=(8k^2-2k)(8k^2-4k+1)=(x+1)x$

one of the two factors (the smaller one ) must =x
and the other factor (bigger) must =x+1 (they are two consecutive positive integers)
and we can judge :
$ 8k^2-4k+1 (smaller)<8k^2-2k (bigger)$
so we set $ x=8k^2-4k+1 $
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
as you said my RHS can be factorized as :
$2k(4k-1)(8k^2-4k+1)=(8k^2-2k)(8k^2-4k+1)=(x+1)x$

one of the two factors (the smaller one ) must =x
and the other factor (bigger) must =x+1 (they are two consecutive positive integers)
and we can judge :
$ 8k^2-4k+1 (smaller)<8k^2-2k (bigger)$
so we set $ x=8k^2-4k+1 $
That would only hold if one of those factors is a prime number.
In general neither is.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
as you said my RHS can be factorized as :
$2k(4k-1)(8k^2-4k+1)=(8k^2-2k)(8k^2-4k+1)=(x+1)x$

one of the two factors (the smaller one ) must =x
and the other factor (bigger) must =x+1 (they are two consecutive positive integers)
and we can judge :
$ 8k^2-4k+1 (smaller)<8k^2-2k (bigger)$
so we set $ x=8k^2-4k+1 $
I think that you are confusing algebraic factorisation with arithmetic factorisation. To take a simple example, if $f(x) = x^2+6x+5$ then $f(x)$ has only one algebraic factorisation, namely $f(x) = (x+1)(x+5)$. If you put $x=1$, then $f(1) = 12$ and the two factors $x+1$ and $x+5$ give you the factorisation $f(1) = 2\times 6$. But if you consider it arithmetically, as a number, then $12$ has other factorisations, such as $12 = 3\times 4$, which are not apparent from the algebraic factorisation of $f(x)$.

For the problem in this thread, we are looking for arithmetic factorisations, and these are much more elusive than the algebraic factorisation suggests.
 

Albert

Well-known member
Jan 25, 2013
1,225
if we set :
$x+1=8k^2-4k+1 , x=8k^2-2k$
then k=0
this does not fit ,because x and k are positive integers

and you said :"That would only hold if one of those factors is a prime number.
In general neither is."

x or x+1 in this case one of them is a prime ,but in other case may be both of them are composite numbers ( for y=8, and y+1=9),anyway they must be coprime

that is :
$ 8k^2-4k+1 , and \,\, 8k^2-2k $ are coprime

using this trait we can find the corresponding value of k and x
 
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