- #1
Albert1
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$a,b\in N$
$k=\dfrac {ab^2-1}{a^2b+1}\,\, \,also \,\,\in N$
find pair(s) of $(a,b)$
$k=\dfrac {ab^2-1}{a^2b+1}\,\, \,also \,\,\in N$
find pair(s) of $(a,b)$
$hint:$Albert said:$a,b\in N$
$k=\dfrac {ab^2-1}{a^2b+1}\,\, \,also \,\,\in N$
find pair(s) of $(a,b)$
The equation is k = (ab^2 - 1) / (a^2b + 1).
The variables are a and b.
Yes, both a and b must be integers and cannot be equal to 0.
One way is to use trial and error by plugging in different values for a and b and checking if the resulting k is an integer. Another way is to use algebraic manipulation to simplify the equation and identify patterns.
No, this equation can only be solved for certain values of k. For example, if k = 0, there are no integer solutions.