Question

For every \(n \in \mathbb{Z}, 4 \nmid\left(n^{2}+2\right)\). Suppose \(a, b \in \mathbb{Z}\). If \(4 \mid\left(a^{2}+b^{2}\right),\) then \(a\) and \(b\) are not both odd.

Short answer

The first statement is true that for every \(n \in \mathbb{Z}, 4 \nmid\left(n^{2}+2\right)\). The second statement is also correct; if \(4 \mid\left(a^{2}+b^{2}\right)\), then \(a\) and \(b\) are not both odd.

Step-by-step solution

Proving the first part of the statement

Considering any integer \(n\) and analyzing the statement \(n^{2}+2\), we can see that if \(n\) is even so \(n^{2}\) as well. Thus, \(n^{2}+2\) will be even, but won't be divisible by 4, since the only even numbers divisible by 4 are those whose last two digits form a number which is multiple of 4, and 2 isn't. On the other hand, if \(n\) is odd, then \(n^{2}\) will be odd, and \(n^{2}+2\) will be even. However, it also won't be divisible by 4 for the same reason stated previously. Thus, it follows that for every \(n \in \mathbb{Z}, 4 \nmid\left(n^{2}+2\right)\).

Proving the second part of the statement

If \(a, b \in \mathbb{Z}\), and \(4 \mid\left(a^{2}+b^{2}\right)\), then \(a\) and \(b\) can't be both odd. If that were the case, then we would have \(a^{2}\) and \(b^{2}\) both being odd (as the square of an odd number is odd), and then their sum \(a^{2}+b^{2}\) would be even. However, for this sum to be divisible by 4, it not only needs to be even but also its last two digits need to form a number which is multiple of 4 - which it isn't, as the sum of two odd numbers gives an even number ending in 0, 2, 4, 6 or 8. Thus, it follows that if \(4 \mid\left(a^{2}+b^{2}\right)\), then \(a\) and \(b\) are not both odd.

Key concepts

Divisibility in Integers

Divisibility is a fundamental concept in number theory, dealing with how one integer can be evenly divided by another. When we say that an integer a is divisible by another integer b, denoted as b | a, it means that when a is divided by b, there is no remainder. For example, 8 is divisible by 2 because 8 divided by 2 equals 4, with no remainder left over.

However, the phrase '4 does not divide n² + 2' is mathematically expressed as 4 ∎ (n² + 2), this indicates that when n² + 2 is divided by 4, there will always be a remainder. To understand this concept using an example, consider the numbers resulting from n² + 2 where n could be any integer. In this case, the numbers that result will never give a remainder of zero when divided by 4, meaning 4 is not a divisor of these numbers.

Proof by Contradiction

Proof by contradiction, also known as indirect proof, is a powerful logical method used to establish the truth of a proposition. The technique involves assuming the opposite of what you aim to prove, and then showing that this assumption leads to a contradiction. When such a contradiction is established, it implies that the original assumption must be false, hence the original proposition is true.

This proof technique is particularly useful in cases where direct proof is complex or elusive. For example, in the proof where we establish that if 4 | (a² + b²), then a and b cannot both be odd, we initially assume that a and b are both odd. If we follow through the logic based on the properties of even and odd numbers, we end up with a statement that contradicts the known properties of numbers. This contradiction implies our original assumption that both a and b were odd must be false, thereby proving the proposition.

Even and Odd Integers

Integers are either even or odd. An even integer is any integer that can be divided by 2 without leaving a remainder, such as 2, 4, or 6. On the other hand, an odd integer is one that, when divided by 2, leaves a remainder of 1, like 1, 3, or 5.

The distinction between even and odd integers is crucial when dealing with divisibility and proofs in mathematics. For instance, the square of an even number is always even, and the square of an odd number is always odd. This characteristic underpins the proof that if 4 | (a² + b²), not both a and b can be odd because the sum of two odd squares would yield an even result, but it would not be divisible by 4, breaking the rules of divisibility and the established properties of even and odd integers.