# find x

#### Albert

##### Well-known member
x and k are positive integers ,satisfying :

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
x and k are positive integers ,satisfying :

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

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!

#### Albert

##### Well-known member

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

what do you think ?

#### Albert

##### Well-known member
$2k(4k-1)(8k^2-4k+1)=(8k^2-2k)(8k^2-4k+1)$

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

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
I don't get it.
How do you know that x does not exist for k > 1?

#### Albert

##### Well-known member
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:
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[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 :

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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
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
$2k(4k-1)(8k^2-4k+1)=(8k^2-2k)(8k^2-4k+1)$

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

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
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
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
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
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
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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