MHB Finding All Possible $k$ for $b,k\in N$

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$b,k\in N$ and $\sqrt {b^2(k-3)(k+1)-4k}$ also $\in N$
find all possible $k$
 
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Albert said:
$b,k\in N$ and $\sqrt {b^2(k-3)(k+1)-4k}$ also $\in N$
find all possible $k$

putting k-2 = a for getting simpler expression we have

$b^2(k-3)(k+1)-4k$
=$b^2(a-2)(a+2)-4(a+1)$
= $b^2(a^2-4)-4(a+1)$
= $a^2b^2 - 4a -4(b^2+1)$
to solve for a
for it to to be perfect square discriminant has to be zero
but 16 + 16b(b^2+1) >0 so no solution
edit: solution is wrong as pointed below
 
Last edited:
kaliprasad said:
putting k-2 = a for getting simpler expression we have

$b^2(k-3)(k+1)-4k$
=$b^2(a-2)(a+2)-4(a+1)$
= $b^2(a^2-4)-4(a+1)$
= $a^2b^2 - 4a -4(b^2+1)$
to solve for a
for it to to be perfect square discriminant has to be zero
but 16 + 16b(b^2+1) >0 so no solution

b = 1, k = 7.
 
greg1313 said:
b = 1, k = 7.
k=4,7
solution k=4 is missed
(b=4)
 
Last edited:
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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