What Are the Possible Values of k Given Specific Conditions?

[Albert1]

(1)$a,b,k\in\mathbb{N}$

(2)$a>b$

(3)$k=\dfrac {a^2+ab+b^2}{ab-1}$

please find :

(i) $\max(k)$

(ii) all possible values of $k$

 

[Opalg]

I am getting nowhere with this, but [sp]there are solutions for $k=4$ (for example $(a,b) = (10,4)$) and $k=7$ ($(a,b) = (2,1)$). I think that those are the only values of $k$ for $k\leqslant 20$, but I do not see how to exclude higher values of $k$. (Headbang)[/sp]

 

[Albert1]

(1)$a,b,k\in\mathbb{N}$

(2)$a>b$

(3)$k=\dfrac {a^2+ab+b^2}{ab-1}$

please find :

(i) $\max(k)$

(ii) all possible values of $k$

$k>\dfrac{\sqrt[3]{3a^3b^3}}{ab}=3---(1)$(AP >GP)
if a,b both are odd numbers then k does not exist
if a,b both are even numbers then k is also a even number (ex:a=4,b=2 then k=4)
so min(k)=4
let a=2m, b=2n,here $m,n \in N$
then :
$k=\dfrac {4(m^2+mn+n^2)}{(4mn-1)}=4x (x\in N)$
for all a,b being even numbers ,we want to prove k=4 then we must prove x=1,that is to prove :
$x=\dfrac{(m^2+mn+n^2)}{(4mn-1)}=1$------(2)(if $x\in N)$
if a odd and b even then k must be odd (ex:a=11 ,b=2 then k=7)
also if a even and b odd then k must be odd
let a=b+d (here b:even and d odd ,so a must be odd)
$k=\dfrac {3b^2+3bd-3+d^2+3}{b^2+bd-1}=3+\dfrac{d^2+3}{b^2+bd-1}$
here $\dfrac{d^2+3}{b^2+bd-1}$ must be even
now we must prove :
$y=\dfrac{d^2+3}{b^2+bd-1}=4$-----(3)(if $y\in N)$

if (2) and (3) can be proved then all is done ,that is min(k)=4 ,and max(k)=7
all the possible valus of k=4 and 7
the proof (2) and (3):I am still thinking ----
By using a program (10001>a>b>0) the possible valus of k=4 and 7
I know (2) and (3) must be true , but how to prove it ?

 

[Opalg]

(1)$a,b,k\in\mathbb{N}$

(2)$a>b$

(3)$k=\dfrac {a^2+ab+b^2}{ab-1}$

My approach was to write this as $a^2+ab+b^2 = k(ab-1)$ and to solve it as a quadratic in $a$, namely $a^2 - b(k-1)a + (b^2+k)$, with solutions \(\displaystyle a = \tfrac12\Bigl(b(k-1) \pm\sqrt{b^2(k-1)^2 - 4(b^2+k)}\Bigr).\) For integer solutions a necessary condition is $b^2(k-1)^2 - 4(b^2+k) = c^2$ for some $c\in\mathbb{N}.$ So we need the Diophantine equation $(k-3)(k-1)b^2 - 4k = c^2$ to have solutions. I could find solutions when $k=4$ or $7$ but not for other values of $k\leqslant20$. That was as far as I could get.

 

[CellCoree]

I would approach this problem by first analyzing the given conditions and finding any restrictions on the values of $k$. From the given information, we can see that $a$, $b$, and $k$ are all natural numbers (positive integers). Additionally, $a$ is greater than $b$.

Next, we can look at the equation for $k$ and see that it involves the variables $a$ and $b$ in both the numerator and denominator. This means that the value of $k$ will change depending on the values of $a$ and $b$.

To find the maximum value of $k$, we can use the given condition that $a>b$. This means that $a$ must be at least one more than $b$, so we can set $a=b+1$. Substituting this into the equation for $k$, we get:
$$k=\dfrac {(b+1)^2+(b+1)b+b^2}{(b+1)b-1}=\dfrac{3b^2+3b+1}{b^2+b-1}$$
To find the maximum value of $k$, we want to maximize the numerator and minimize the denominator. The maximum value of the numerator will occur when $b$ is the largest possible value, which is when $b=1$. The minimum value of the denominator will occur when $b$ is the smallest possible value, which is when $b=2$. Substituting these values into the equation for $k$, we get:
$$\max(k)=\dfrac{3(1)^2+3(1)+1}{(1)^2+(1)-1}=\dfrac{7}{1}=7$$

To find all possible values of $k$, we can start by listing out some values for $a$ and $b$ that satisfy the given conditions. For example, we could have $a=3$ and $b=2$, which gives us:
$$k=\dfrac {3^2+3(2)+(2)^2}{(3)(2)-1}=\dfrac{13}{5}$$
We could also have $a=4$ and $b=3$, which gives us:
$$k=\dfrac {4^2+4(3)+(3)^2}{(4)(3)-1}=\dfrac{25}{