I am a proper fraction.
The sum of my numerator and my denominator is 1 less than a perfect square.
The difference of my denominator and my numerator is 1 more than a perfect square.
The product of my numerator and my denominator is 1 less than a perfect square.
What fraction am I?
To find the fraction that satisfies the given conditions, we can represent the fraction as "x/y" where x is the numerator and y is the denominator.
According to the conditions:
1) The sum of the numerator and the denominator is 1 less than a perfect square.
So, we can write: x + y = n^2 - 1, where n is a positive integer.
2) The difference between the denominator and the numerator is 1 more than a perfect square.
So, we can write: y - x = m^2 + 1, where m is a positive integer.
3) The product of the numerator and denominator is 1 less than a perfect square.
So, we can write: x * y = p^2 - 1, where p is a positive integer.
We can use these equations to solve for x and y.
From the first equation, we can rearrange it to solve for x:
x = (n^2 - 1) - y
Substituting x into the second equation, we have:
y - ((n^2 - 1) - y) = m^2 + 1
2y - n^2 + 1 = m^2 + 1
2y = m^2 + n^2
y = (m^2 + n^2) / 2
Substituting the values of x and y into the third equation, we get:
((n^2 - 1) - y) * y = p^2 - 1
(n^2 - 1) * y - y^2 = p^2 - 1
(n^2 - 1) * ((m^2 + n^2) / 2) - ((m^2 + n^2) / 2)^2 = p^2 - 1
Simplifying the equation, we have:
(n^2 - 1) * (m^2 + n^2) - ((m^2 + n^2)^2) / 4 = p^2 - 1
Since we are looking for a proper fraction, both x and y have to be positive integers. This implies that n and m are positive integers as well.
We can solve this equation and substitute different values of n and m to find the valid fraction that satisfies all the conditions.