MHB What Are the Possible Values of k Given Specific Conditions?

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The discussion focuses on determining the possible values of k given the equation k = (a^2 + ab + b^2) / (ab - 1) under the conditions that a, b, and k are natural numbers with a > b. The participants identify that k can take values of 4 and 7, with examples provided for each case. They explore the conditions under which k is even or odd, concluding that k must be at least 4 and can reach a maximum of 7. Attempts to prove these bounds involve analyzing quadratic equations and Diophantine equations, but participants express difficulty in excluding higher values of k. Ultimately, the consensus is that the only possible values of k are 4 and 7.
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(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 $k$
 
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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]
 
Albert said:
(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 ?:confused:
 
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Albert said:
(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 $$a = \tfrac12\Bigl(b(k-1) \pm\sqrt{b^2(k-1)^2 - 4(b^2+k)}\Bigr).$$ 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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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