MHB Solving for x and k: Positive Ints

[Albert1]

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

 

[I like Serena]

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)

 

[Albert1]

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 ?

 

[Albert1]

$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 $

 

[I like Serena]

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

 

[Albert1]

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 :

[FONT=MathJax_Main]2[FONT=MathJax_Math]k[FONT=MathJax_Main]([FONT=MathJax_Main]4[FONT=MathJax_Math]k[FONT=MathJax_Main]−[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Main]([FONT=MathJax_Main]8[FONT=MathJax_Math]k[FONT=MathJax_Main]2[FONT=MathJax_Main]−[FONT=MathJax_Main]4[FONT=MathJax_Math]k[FONT=MathJax_Main]+[FONT=MathJax_Main]1[FONT=MathJax_Main])[FONT=MathJax_Main]=[FONT=MathJax_Main]([FONT=MathJax_Main]8[FONT=MathJax_Math]k[FONT=MathJax_Main]2[FONT=MathJax_Main]−[FONT=MathJax_Main]2[FONT=MathJax_Math]k[FONT=MathJax_Main])[FONT=MathJax_Main]([FONT=MathJax_Main]8[FONT=MathJax_Math]k[FONT=MathJax_Main]2[FONT=MathJax_Main]−[FONT=MathJax_Main]4[FONT=MathJax_Math]k[FONT=MathJax_Main]+[FONT=MathJax_Main]1[FONT=MathJax_Main])

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

let :

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

we have :

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

[FONT=MathJax_AMS]∴[FONT=MathJax_Math]k[FONT=MathJax_Main]=[FONT=MathJax_Main]1[FONT=MathJax_Main],[FONT=MathJax_Math]x[FONT=MathJax_Main]=[FONT=MathJax_Main]5

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

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

 

[I like Serena]

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]

$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).

 

[I like Serena]

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.

 

[Albert1]

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 like Serena]

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]

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.

 

[Albert1]

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