Bunuel
If \(a^2+b^2=c^2\), and a, b, and c are all integers. Which of the following CANNOT be the value of a+b+c?
A. -2
B. 1
C. 2
D. 4
E. 6
Since \(a^2+b^2=c^2\), we can have the following cases for a, b and c considering whether they are even or odd:
Case 1: a & b are even => c is even => a + b + c = even
Case 2: a is even & b is odd (or vice-versa) => c is odd => a + b + c = even
Case 3: a & b are odd => This case is not possible since sum of squares of 2 odd numbers would be an even number that is only a multiple of 2; NOT a multiple of 4 (ex. \(3^2+5^2=34\) which is not a multiple of 4); hence it cannot be a square number
[Note - even if you did not realize this, you would have assumed a and b are odd and c is even => a + b + c = even - just as in the previous cases]
Thus, the value of \(a+b+c\) is always even [Note: a, b, c can be negative; hence, the sum can be a negative value as well]
Thus, the sum cannot be 1 (an odd number)
Alternative approach:Since the question doesn't talk about positive integers (only integers), we can easily take a = 0 and b as a suitable value to verify the options:
Option A: a = 0, b = -1 => \(c^2\) = 1 => c = -1 => a + b + c = -2
Option C: a = 0, b = 1 => \(c^2\) = 1 => c = 1 => a + b + c = 2
Option D: a = 0, b = 2 => \(c^2\) = 4 => c = 2 => a + b + c = 4
Option E: a = 0, b = 3 => \(c^2\) = 9 => c = 3 => a + b + c = 6
Only for Option B, we do not have a suitable combination of integer values
Answer BLearn about number properties:
1. Numbers 1:
https://youtu.be/UkG3pa1Mkis2. Numbers 2:
https://youtu.be/G-uthVRTy9Q3. Numbers 3:
https://youtu.be/K1jMgFJ5qdI