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

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Discussion Overview

The discussion centers around finding all possible natural number values of \( k \) given the condition that \( b, k \in \mathbb{N} \) and that the expression \( \sqrt{b^2(k-3)(k+1)-4k} \) is also a natural number. The scope includes mathematical reasoning and exploration of conditions under which the expression holds true.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Some participants propose analyzing the expression \( b^2(k-3)(k+1)-4k \) to determine under what conditions it can yield a perfect square.
  • Others suggest substituting specific values for \( b \) to simplify the exploration of possible \( k \) values.
  • A later reply questions whether there are restrictions on \( k \) based on the requirement that the entire expression under the square root must remain non-negative.
  • Some participants discuss the implications of \( k \) being a natural number and how that affects the factors \( (k-3) \) and \( (k+1) \).

Areas of Agreement / Disagreement

Participants have not reached a consensus on the values of \( k \) that satisfy the given conditions, and multiple competing views on the approach to solving the problem remain.

Albert1
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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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