The quantity $(x+m)^2- (x+n)^2 = (m-n)^2$, where $m$ and $n$ are unequal non zero constants, is satisfied by $x = am +bn$, where:
- $a = 0, b$ has a unique non-zero value
- $a = 0, b$ has two non-zero values
- $b = 0, a$ has a unique non-zero values
- $b = 0, a$ has two non-zero values
- $a$ and $b$ each have a unique non-zero value.
Expand the equation: (x^2 + 2xm + m^2) - (x^2 + 2xn + n^2) = (m-n)^2. This simplifies to 2x(m-n) + m^2 - n^2 = m^2 - 2mn + n^2. Further: 2x(m-n) = 2n^2 - 2mn = -2n(m-n). Since m is not equal to n, we can divide by 2(m-n) to get x = -n. This matches x = am + bn with a = 0 and b = -1.
Applying the difference of squares identity to the left side, we rewrite the equation as (x + m + x + n)(x + m - x - n) = (m - n)^2. This simplifies to (2x + m + n)(m - n) = (m - n)^2, and dividing both sides by the non-zero term (m - n) gives 2x + m + n = m - n. Solving for x yields 2x = -2n, which means x = -n. Substituting this result into the given form x = am + bn, we get -n = am + bn, which implies am + (b+1)n = 0. Because m and n are independent, non-zero constants, their coefficients must be zero, resulting in a = 0 and b = -1, which is a unique non-zero value. Therefore, a equals 0 and b has a unique non-zero value.