Question

Show that if \(p\) is a prime and \(p \equiv 1(\bmod 4)\), then there is an integer \(x\) such that \(x^{2} \equiv-1(\bmod p)\).

Short answer

Question: Prove that for a prime number \(p\) satisfying \(p \equiv 1(\bmod 4)\), an integer \(x\) exists such that \(x^2 \equiv -1(\bmod p)\). Answer: We can prove this by following these steps: 1. Use Fermat's Little Theorem to rewrite the condition as \(a^{4k} \equiv 1(\bmod p)\). 2. Choose a non-quadratic residue modulo \(p\) and call it \(a\). 3. Apply Fermat's Little Theorem to the non-square element, \(a^{4k} \equiv 1(\bmod p)\). 4. Define \(x \equiv a^{2k}(\bmod p)\), and then show that \(x^2 \equiv (x + 1)(x - 1) \equiv -1(\bmod p)\). In conclusion, we can prove that for any prime number \(p\) satisfying \(p \equiv 1(\bmod 4)\), there exists an integer \(x\) such that \(x^2 \equiv -1(\bmod p)\).

Step-by-step solution

Fermat's Little Theorem

Recall that Fermat's Little Theorem states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then \(a^{p-1} \equiv 1(\bmod p)\). In this problem, \(p \equiv 1(\bmod 4)\), so we can write \(p = 4k+1\) for some integer \(k\). Then, we can rewrite Fermat's Little Theorem as \(a^{4k} \equiv 1(\bmod p)\).

Consider a non-square element

Choose an element \(a\) that is not a quadratic residue (i.e., a value that cannot be expressed as a square of any integer) modulo \(p\). We know that such an element exists because, by the Quadratic Reciprocity theorem, there are \(\frac{p-1}{2}\) quadratic residues and \(\frac{p-1}{2}\) non-residues in the set \(\{1, 2, ..., p-1\}\). Let \(a\) be a non-quadratic residue modulo \(p\).

Using Fermat's Little Theorem on the non-square element

Now, raise \(a\) to the power of \((p-1)/2\), which is equal to \(a^{4k}\). By Fermat's Little Theorem, we have \(a^{4k} \equiv 1(\bmod p)\). But since \(p = 4k+1\), we can rewrite this as \(a^{2(2k)} \equiv 1(\bmod p)\).

Define \(x\) and check the condition

Now define \(x \equiv a^{2k}(\bmod p)\). To find \(x^2\), we need to square both sides and reduce modulo \(p\). This gives us: \(x^2 \equiv a^{4k}(\bmod p)\). From step 3, we know that \(a^{4k} \equiv 1(\bmod p)\). Therefore, we can substitute this result into the equation: \(x^2 \equiv 1(\bmod p)\). Since \(a\) is a non-quadratic residue modulo \(p\), it follows that \(x \not\equiv \pm 1(\bmod p)\). Hence, \(x^2 - 1 \not\equiv 0(\bmod p)\), and therefore \(p\) must divide the difference \((x^2 - 1)\). To show this, we can rewrite \(x^2 - 1\) as a product of factors: \(x^2 - 1 = (x - 1)(x + 1)\). Since \(p\) divides the product and \(p\) is a prime, it can divide either \(x-1\) or \(x+1\) but not both since their difference is 2, and \(p > 2\). Without loss of generality, let's assume that \(p\) divides \(x - 1\): \(x - 1 \equiv 0(\bmod p)\). Now we can solve for \(x\): \(x \equiv 1(\bmod p)\). Finally, using the value of \(x\), we can check the expression \(x^2 \equiv -1(\bmod p)\): \(x^2 \equiv (x + 1)(x - 1) \equiv -1(\bmod p)\). Thus, we have shown that there is an integer \(x\) such that \(x^2 \equiv -1(\bmod p)\) for any prime number \(p\) satisfying \(p \equiv 1(\bmod 4)\).

Key concepts

Fermat's Little Theorem

When it comes to working with prime numbers and modular arithmetic, Fermat's Little Theorem is incredibly useful. It provides us with a way to calculate powers of integers under modulo operations, significantly simplifying many complex calculations. The theorem states that if you have a prime number, let's say \(p\), and an integer \(a\) which is not divisible by \(p\), then raising \(a\) to the power of \(p-1\) and taking the modulus of \(p\) equals one: \(a^{p-1} \bmod p = 1\).

This theorem underpins a vast array of mathematical concepts and is frequently used to demonstrate the existence of certain numbers within modular systems, such as the existence of an integer \(x\) for which \(x^2 \bmod p = -1\), given certain conditions on \(p\).

Application of Fermat's Little Theorem to prime numbers modulo 4:

When a prime number \(p\) is one more than a multiple of four (i.e., \(p \bmod 4 = 1\)), Fermat's Little Theorem can be leveraged to show that \(p\) will have particular residues when squared and taken under modulus operations. This attribute of prime numbers allows us to find integers that exhibit specific modular properties, which is where the connection with the problem at hand becomes apparent.

Quadratic Residue

The concept of a quadratic residue is at the heart of many problems in number theory. In simple terms, a quadratic residue modulo \(p\) is an integer \(x\) such that there exists some integer \(a\) where \(x \bmod p = a^2 \bmod p\). That is, \(x\) can be expressed as the square of some integer modulo \(p\). Conversely, a non-quadratic residue cannot be expressed this way, which means no square equals that number when taken modulo \(p\).

This distinction is critical when working with equations that involve squares in modular arithmetic, as it determines whether solutions exist or not.

In the context of the exercise, by choosing an element that is a non-quadratic residue, we ensure that when we square it and apply modulo \(p\), it won't simply roll back to the original number or its negative equivalent. This strategy is used to cleverly navigate the terrain of modular arithmetic and find an integer \(x\) that satisfies the condition \(x^2 \bmod p = -1\), which might not be immediately apparent.

Quadratic Reciprocity

Quadratic reciprocity is a profound result in number theory that allows us to understand the relationship between quadratic residues and non-residues for different primes. Essentially, it tells us that except for some well-understood exceptions, the solvability of the quadratic equation \(x^2 \bmod p\) is reciprocal for two odd primes: knowing whether a number is a residue modulo one prime gives us insight into whether it's a residue modulo another. This concept is intriguing because it links the properties of numbers across different modular systems.

The theorem provides insights not only into individual numbers and their properties when subjected to different primes but also adds depth to our understanding of the structure of numbers under modular arithmetic as a whole.

By employing quadratic reciprocity in our exercise, we can assess the number of quadratic residues and non-residues, as the theorem assures us of an equal distribution among the integers from 1 to \(p - 1\), where \(p\) is a prime number. This balance between residues and non-residues is what permits the strategy used in the step-by-step solution, guiding us towards demonstrating the existence of an integer that satisfies the given modular equation.