Admitting the digit 0 as a square the problem is quite easy.
Consider a square n=m^2
Substracting 2m-1 and adding an square a^2 (just taking care that a^2 is even and its value is one less than 2m-1), the problem is done.
2m-1=a^2+1, so m=(a^2+2)/2 and we have the second number, composed by the squares of m-1 and a. This is true for each value of a (a is even).
The third number is always n+1.
Ex (to begin assign a value to "a"):
a=14 m=99 n=9801
But if 0 is not admitted as a square I don't know if the ennunciate is true.
Edited on February 11, 2016, 4:32 am
Posted by armando
on 2016-02-11 04:15:29