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

[Albert1]

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

 

[kaliprasad]

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

Spoiler

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

 

[Greg]

Spoiler

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

Spoiler

b = 1, k = 7.

 

[Albert1]

Spoiler

b = 1, k = 7.

Spoiler

k=4,7
solution k=4 is missed
(b=4)