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

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The discussion focuses on identifying all possible natural numbers \( k \) such that both \( b \) and \( k \) are natural numbers, and the expression \( \sqrt{b^2(k-3)(k+1)-4k} \) is also a natural number. The participants explore the conditions under which the expression remains a perfect square, leading to specific values of \( k \). The conclusion indicates that valid values of \( k \) must satisfy the derived conditions from the quadratic expression formed by the equation.

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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:

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