Find Integer Solutions to $k=\dfrac{ab^2-1}{a^2b+1}$

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In summary, the equation for finding integer solutions to this problem is k = (ab^2 - 1) / (a^2b + 1). The variables in this equation are a and b. There are restrictions on the values of a and b as they must be integers and cannot be equal to 0. To find integer solutions, one can use trial and error or algebraic manipulation. However, this equation cannot be solved for all integer values of k, as some values such as 0 do not have any integer solutions.
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Albert1
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$a,b\in N$

$k=\dfrac {ab^2-1}{a^2b+1}\,\, \,also \,\,\in N$

find pair(s) of $(a,b)$
 
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Albert said:
$a,b\in N$

$k=\dfrac {ab^2-1}{a^2b+1}\,\, \,also \,\,\in N$

find pair(s) of $(a,b)$
$hint:$
$if \,\,a=1\,\, then \,\,b=?$
$if \,\,a>1\,\, then \,\,no \,\,solution.\,\, why ?$
 

FAQ: Find Integer Solutions to $k=\dfrac{ab^2-1}{a^2b+1}$

What is the equation for finding integer solutions to this problem?

The equation is k = (ab^2 - 1) / (a^2b + 1).

What are the variables in this equation?

The variables are a and b.

Are there any restrictions on the values of a and b?

Yes, both a and b must be integers and cannot be equal to 0.

How can you find integer solutions to this equation?

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.

Can this equation be solved for all integer values of k?

No, this equation can only be solved for certain values of k. For example, if k = 0, there are no integer solutions.

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