Edward Frenkel --- $\textbf{Exercise 8.}$ Find two distinct square roots of $i$. $\textbf{Solution 8.}$ Suppose $a$ and $b$ are real numbers such that $(a + bi)^2 = i.$ Then $i = (a + bi)^2 = (a^2 - b^2) + 2abi.$ Thus $a^2 = b^2 \text{ and } ab = \frac{1}{2}.$ The equation $a^2 = b^2$ implies that $a = b$ or $a = -b$. However, if $a = -b$, the equation $ab = \frac{1}{2}$ implies that $-b^2 = \frac{1}{2}$, which is impossible because $b$ is a real number. Thus we have $a = b$. The equation $ab = \frac{1}{2}$ now becomes the equation $b^2 = \frac{1}{2}$, which leads to $a = b = \pm\frac{\sqrt{2}}{2}$. Hence the only two possibilities for square roots of $i$ are $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \text{ and } -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i.$ Squaring each of the numbers above gives $i$. Thus the two numbers above are indeed square roots of $i$. $\textit{Commentary:}$ This exercise asks to find two different complex numbers whose square is equal to $i$. It illustrates that, unlike in the real numbers, quadratic equations in the complex numbers can have multiple distinct solutions. Geometrically, the two square roots of $i$ are reflections of each other across the line $y=x$ in the complex plane. They lie on the unit circle, halfway between $1$ and $i$, and halfway between $-1$ and $-i$. Finding square roots in the complex numbers is a fundamental operation with applications in various areas of mathematics, including the solution of quadratic equations, the calculation of principal values of multivalued functions (such as the logarithm and the square root itself), and the study of Riemann surfaces. $\textit{Example:}$ The two square roots of $-1$ in the complex numbers are $i$ and $-i$, because: $i^2 = (-i)^2 = -1$ In $\mathbb{R}$, the square root of a positive real number $a$ is the unique positive real number $b$ such that $b^2 = a$. The square root of a negative real number does not exist in $\mathbb{R}$. --- In $\mathbb{F}_p$ for prime $p$, an element $a$ has a square root if and only if $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$. For example, in $\mathbb{F}_7$, 2 has a square root (namely, 3 and 4) because $2^3 \equiv 1 \pmod{7}$, but 5 does not have a square root because $5^3 \equiv -1 \pmod{7}$. $3^2 \equiv 9 \equiv 2 \pmod{7}$ $4^2 \equiv 16 \equiv 2 \pmod{7}$ Explanation: The statement you provided is correct, but there seems to be a misunderstanding about the square root of 2 modulo 7. Let me clarify: In $\mathbb{F}_p$ for prime $p$, an element $a$ has a square root if and only if $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$. This is known as Euler's criterion. In the example of $\mathbb{F}_7$: 1) For $a = 2$, we have: $2^{\frac{7-1}{2}} \equiv 2^3 \equiv 8 \equiv 1 \pmod{7}$ This means that 2 has a square root in $\mathbb{F}_7$. 2) For $a = 5$, we have: $5^{\frac{7-1}{2}} \equiv 5^3 \equiv 125 \equiv -1 \pmod{7}$ This means that 5 does not have a square root in $\mathbb{F}_7$. Now, regarding the square root of 2 in $\mathbb{F}_7$: The square roots of 2 in $\mathbb{F}_7$ are actually 3 and 4, because: $3^2 \equiv 9 \equiv 2 \pmod{7}$ $4^2 \equiv 16 \equiv 2 \pmod{7}$ So, both 3 and 4 are square roots of 2 in $\mathbb{F}_7$. In general, if an element has a square root in $\mathbb{F}_p$, it will have two distinct square roots (unless the element is 0 or the element is $\frac{p-1}{2}$ when $p \equiv 1 \pmod{4}$). --- In the field of complex numbers, every number has two square roots (counting multiplicity). This is a consequence of the Fundamental Theorem of Algebra, which states that every nonconstant polynomial with complex coefficients has a root in the complex numbers. Applying this to the polynomial $z^2 - a$ for any complex number $a$ yields the existence of a square root of $a$. The process of finding square roots in the complex numbers can be generalized to finding $n$-th roots for any positive integer $n$. The $n$-th roots of a complex number $z$ are the solutions to the equation $w^n = z$. There are always $n$ such roots (counting multiplicity), and they are evenly spaced around a circle centered at the origin in the complex plane. --- ## Finite Fields In the finite field $\mathbb{F}_7$, there are only two square roots of 2: 3 and 4. This is because if we square each element of the field, we get: $0^2 \equiv 0 \pmod{7}$ $1^2 \equiv 1 \pmod{7}$ $2^2 \equiv 4 \pmod{7}$ $3^2 \equiv 2 \pmod{7}$ $4^2 \equiv 2 \pmod{7}$ $5^2 \equiv 4 \pmod{7}$ $6^2 \equiv 1 \pmod{7}$ As you can see, only 3 and 4 square to 2 in this field. Now, let's look at some more examples of the square root of 2 in other finite fields: 1) In $\mathbb{F}_{11}$, the square roots of 2 are 6 and 5: $6^2 \equiv 36 \equiv 2 \pmod{11}$: seems wrong, should be 3 $5^2 \equiv 25 \equiv 2 \pmod{11}$:seems wrong, should be 3 --- You are absolutely right, and I apologize for my mistake. Let me correct it: In $\mathbb{F}_{11}$, the square roots of 3 (not 2) are 5 and 6. $5^2 \equiv 25 \equiv 3 \pmod{11}$ $6^2 \equiv 36 \equiv 3 \pmod{11}$ To verify, we can check the squares of all elements in $\mathbb{F}_{11}$: $0^2 \equiv 0 \pmod{11}$ $1^2 \equiv 1 \pmod{11}$ $2^2 \equiv 4 \pmod{11}$ $3^2 \equiv 9 \pmod{11}$ $4^2 \equiv 5 \pmod{11}$ $5^2 \equiv 3 \pmod{11}$ $6^2 \equiv 3 \pmod{11}$ $7^2 \equiv 5 \pmod{11}$ $8^2 \equiv 9 \pmod{11}$ $9^2 \equiv 4 \pmod{11}$ $10^2 \equiv 1 \pmod{11}$ As we can see, 2 does not have a square root in $\mathbb{F}_{11}$, but 3 has two square roots: 5 and 6. Thank you for pointing out the error. It's important to be precise when discussing mathematical concepts. --- 2) In $\mathbb{F}_{13}$, 2 does not have a square root. We can check this using Euler's criterion: $2^{\frac{13-1}{2}} \equiv 2^6 \equiv 64 \equiv -1 \pmod{13}$ : Since this is not congruent to 1, 2 does not have a square root in $\mathbb{F}_{13}$. 3) In $\mathbb{F}_{17}$, the square roots of 2 are 6 and 11: $6^2 \equiv 36 \equiv 2 \pmod{17}$ $11^2 \equiv 121 \equiv 2 \pmod{17}$ 4) In $\mathbb{F}_{19}$, the square roots of 2 are 5 and 14: $5^2 \equiv 25 \equiv 2 \pmod{19}$ $14^2 \equiv 196 \equiv 2 \pmod{19}$ In general, if 2 has a square root in a finite field $\mathbb{F}_p$ (where $p$ is an odd prime), it will have exactly two square roots. If 2 doesn't have a square root, Euler's criterion will always give -1.