2026/090) Let a, b, c be the real numbers. It is true, that a+b, b+c and c+a are three consecutive integers, in a certain order, and the smallest of them is odd. Prove that the numbers a, b, c are also consecutive integers in a certain order.

 

The expressions are cyclic so WLOG we can assume

a +b < a +c < b+c

From given conditions

$a+b = 2n -1\cdots(1)$ as the lowest number is odd

$a+c = 2n\cdots(2)$

$b+c = 2n +1\cdots(3)$

Adding these 3 we get

$2a + 2b + 2c = 6n$

or $a +b +c = 3n\cdots(4)$

so $a = (a +b + c)-(b+c) = 3n -(2n+1) = n-1$

$b= a+b -a = (2n)-(2n-1) = n$

$c = a+c -a = n+1$

so all are consecutive integers