Question

Prove the statements below. You can prove these by considering remainders modulo n

(a) For all integers a, b, a^{2}+b^{2}+1 is not divisible by 4 .
(b) For all integer

Answer 1

To prove statement (a), we consider different cases for the parity of a and b. If they are both even, a^2+b^2 is divisible by 4, but adding 1 makes it not divisible by 4. If a is odd and b is even (or vice versa), a^2+b^2 is also not divisible by 4. Adding 1 again makes it not divisible by 4. If both a and b are odd, a^2+b^2 is divisible by 2, but not by 4. Adding 1 makes it not divisible by 4. To prove statement (b), we use the fact that the sum of the remainders when a number is divided by 3 is congruent to the remainder when the sum of its digits is divided by 3. Since a, a^2, and 1 all have remainders of 1 when divided by 3, their sum will also have a remainder of 1 when divided by 3. Therefore, a^2+a+1 is not divisible by 3.

Step-by-step explanation:

(a) For all integers a, b, a2+b2+1 is not divisible by 4:

To prove this statement, we'll consider different cases:

• If a and b are both even, then a2+b2 is divisible by 4, but adding 1 makes it not divisible by 4.
• If a is odd and b is even (or vice versa), then a2+b2 is also not divisible by 4. Adding 1 again makes it not divisible by 4.
• If both a and b are odd, then a2+b2 is divisible by 2, but not by 4. Adding 1 makes it not divisible by 4.

(b) For all integers a, a2+a+1 is not divisible by 3:

To prove this statement, we can use the fact that the sum of the remainders when a number is divided by 3 is congruent to the remainder when the sum of its digits is divided by 3. Since a, a2, and 1 all have remainders of 1 when divided by 3, their sum will also have a remainder of 1 when divided by 3. Therefore, a2+a+1 is not divisible by 3.