That is a very long way to Rome as well. Since the original equations are symmetric, we may assume that the outcome will be something symmetric, too. Thus, when modelling x and y by a symmetric expression (x = a+b, y=a-b), we can drastically shorten the road. We land at the same quadratic equation, but without any intermediate jungles of terms. Already in x^5+y^5, the uneven powers of b cancel out, thus leaving the quadratic immediately. However, it's always interesting to see the different roads you could go if you are up to, landing always at the same result.
That was a lot of fun! Clever problem and clever solution - thank you!
That is a very long way to Rome as well.
Since the original equations are symmetric, we may assume that the outcome will be something symmetric, too. Thus, when modelling x and y by a symmetric expression (x = a+b, y=a-b), we can drastically shorten the road. We land at the same quadratic equation, but without any intermediate jungles of terms. Already in x^5+y^5, the uneven powers of b cancel out, thus leaving the quadratic immediately.
However, it's always interesting to see the different roads you could go if you are up to, landing always at the same result.
SENSATIONAL PROFESSOR!😊
Nicely explained
👏👏👏