hasty pulsar
❗️Guys, please check my solution . Point out the errors and tell me how to write a perfect solution. * It is necessary to find possible pairs of integer values for m,n ❗️
Add mn to both sides:
The left-hand side is a perfect square (sum of squares) => the right-hand side mn(mn+1) is also a square number (from theory).
The GCD of mn and mn+1 is 1 => the numbers are coprime.
If the numbers are coprime, and their product is a square => mn and mn+1 must also be squares (This is easy to prove: if the product has an even degree for each prime factor, and mn and mn+1 do not divide each other (GCD = 1 => prime numbers), then each of them must contain its own degree — also even, because their product is a square).
Two squares: mn and mn+1 differ by one => these are two consecutive squares, and this only happens when: x^2 - y^2 = 1
x^2 = mn+1
y^2 = mn
The only squares that differ by one are 0 and 1. Why?
We rewrite the difference differently: (x - y)(x + y) = 1.
1 must be represented as a product of two integers:
1.) -1 \cdot (-1) = 1
2.) 1 \cdot 1 = 1
We write the system and solve for each case (substitution method):
1.) Case:
x - y = -1
x + y = -1
Adding these: 2x = -2
x = -1
y = 0
2.) Case:
x - y = 1
x + y = 1
Adding these: 2x = 2
x = 1
y = 0
It follows that x = \pm 1, y = 0 => mn = 0 (y^2 = 0^2 = 0) => mn+1 = 1.
If mn = 0 => mn(mn+1) = 0.
*It should be noted that for the equation to hold, mn can not only be zero but also equal to -1, since in that case, the second factor mn + 1 becomes zero => mn \cdot 0 = 0.
We write this in another form: (m + n)^2 = 0 => m + n = 0.
lusty bobcat
velvet garnet