Grade 11/12 Math Circles
Cryptography, Part 2 - Solutions

Exercise Solutions

Problem Set Solutions

1. Which of the following numbers are prime? For the ones that aren’t prime, factor them as a product of primes.

   1. 101

   2. 119

   3. 127

   Solution:

   1. The number $101$ is prime. You can verify this using trial division – if $101$ were not prime, it would have a prime factor smaller than $11$. (Can you explain why?) But $101$ is not divisible by any primes smaller than $11$, so it must be prime.

   2. The number $119$ is not prime, because it factors as $119=7\cdot 17$.

   3. The number $127$ is prime. Just as you can argue in part (a), if this number were not prime, it would have a prime factor smaller than $13$. But $127$ is not divisible by any primes smaller than $13$.

2. 1. For any positive integer $a$, what is $gcd(a,0)$ equal to?

   2. Calculate $gcd(15,27)$ using the Euclidean algorithm.

   3. Calculate $gcd(61783,4019881)$ using the Euclidean algorithm.

   Solution:

   1. For any positive integer $a$, $gcd(a,0)=a$. Indeed, $a$ divides itself, and $a$ divides $0$ because $0=a\cdot 0$. Any integer larger than $a$ cannot divide $a$, so $a$ is indeed the greatest common divisor of $a$ and $0$.

   2. We perform divisions with remainder, starting with dividing $27$ by $15$. $$\begin{array}{rl}27& =1\cdot 15+12\\ 15& =1\cdot 12+3\\ 12& =4\cdot 3+0.\end{array}$$ We see that $3$ is the last non-zero remainder, so $gcd(15,27)=3$.

   3. Again, we perform divisions with remainder, starting with dividing $4019881$ by $61783$. $$\begin{array}{rl}4019881& =65\cdot 61783+3986\\ 61783& =15\cdot 3986+1993\\ 3986& =2\cdot 1993+0.\end{array}$$ We see that $1993$ is the last non-zero remainder, so $gcd(61783,4019881)=1993$.

3. Find integers $x$ and $y$ solving the following equations.

   1. $15x+27y=3$. (Hint: We’ve done half the work in the previous question.)

   2. $17x+31y=1$.

   3. $17x+31y=4$. (Hint: This requires a little more than just the Extended Euclidean algorithm. How can you work with the answer obtained in part (b) to get an answer here?)

   Solution:

   1. Starting from the Euclidean algorithm work from question 2(b), we have $$\begin{array}{rl}27& =1\cdot 15+12\\ 15& =1\cdot 12+3\\ 12& =4\cdot 3+0.\end{array}$$ Leaving out the last line, solving for the remainders, and writing them in reverse order, we get $$\begin{array}{rl}3& =15-1\cdot 12\\ 12& =27-1\cdot 15.\end{array}$$ Substituting the second equation into the first, we get $$\begin{array}{rl}3& =15-1\cdot 12\\ 3& =15-1\cdot (27-1\cdot 15)\\ 3& =2\cdot 15-1\cdot 27.\end{array}$$ This gives us the solution $x=2$ and $y=-1$ to the equation $15x+27y=3$.

   2. Here, we first have to run the Euclidean algorithm on $17$ and $31$, which gives us $$\begin{array}{rl}31& =1\cdot 17+14\\ 17& =1\cdot 14+3\\ 14& =4\cdot 3+2\\ 3& =1\cdot 2+1\\ 2& =2\cdot 1+0.\end{array}$$ Removing the last line, solving for the remainders, and writing them in reverse order, we get $$\begin{array}{rl}1& =3-1\cdot 2\\ 2& =14-4\cdot 3\\ 3& =17-1\cdot 14\\ 14& =31-1\cdot 17.\end{array}$$ Substituting each line one at a time, we get $$\begin{array}{rl}1& =3-1\cdot 2\\ & =3-1\cdot (14-4\cdot 3)\\ & =3-1\cdot 14+4\cdot 3\\ & =5\cdot 3-1\cdot 14\\ & =5\cdot (17-1\cdot 14)-1\cdot 14\\ & =5\cdot 17-5\cdot 14-1\cdot 14\\ & =5\cdot 17-6\cdot 14\\ & =5\cdot 17-6\cdot (31-1\cdot 17)\\ & =5\cdot 17-6\cdot 31+6\cdot 17\\ & =11\cdot 17-6\cdot 31.\end{array}$$ This gives us the solution $x=11$ and $y=-6$ to the equation $17x+31y=1$.

   3. Given our work in part (b), there is just one more thing to do. We already know that $17\cdot 11+31\cdot (-6)=1$, and if we multiply both sides by $4$, we get $$17\cdot (11\cdot 4)+31\cdot (-6\cdot 4)=4.$$ This gives a solution $x=11\cdot 4=44$ and $y=-6\cdot 4=-24$ to the equation $17x+31y=4$.

4. Compute the following values of the Euler phi function.

   1. $\varphi (7)$

   2. $\varphi (15)$

   3. $\varphi (27)$

   4. $\varphi (30)$

   5. $\varphi (p^k)$, where $p$ is a prime number and $k$ is a positive integer.

   Solution:

   1. Since $7$ is prime, we apply the formula for $\varphi (p)$ where $p$ is prime from Example 5. This gives us $\varphi (7)=7-1=6$.

   2. Since $15=3\cdot 5$ is the product of two different primes, we can use the formula for $\varphi (n)$ in the case discussed in Example 6. This gives us $\varphi (15)=(3-1)(5-1)=2\cdot 4=8$.

   3. Since $27=3^3$, we haven’t discussed a formula for $\varphi (n)$ that will work here. We could try all the numbers between $1$ and $27$ and take gcds, but let’s be a little bit more clever.

      Since $3$ is the only prime dividing $27$, we get $gcd(m,27)\ne 1$ exactly when $m$ is a multiple of $3$. So, we just have to count the number of multiples of $3$ between $0$ and $26$ and subtract them from $27$ to get the answer. Now, there are exactly nine multiples of $3$ in that range: $$3,6,9,12,15,18,21,24,27.$$ Hence $\varphi (27)=27-9=18$.

   4. Notice that $30=2\cdot 3\cdot 5$, so we haven’t discussed any tricks for calculating $\varphi (30)$ that will compute it instantly. But again, given an integer $m$ between $1$ and $30$, we will have $gcd(m,30)\ne 1$ exactly when $m$ is a multiple of either $2,3$, or $5$. So, let’s start by removing all multiples of $2$ from the list of integers from $1$ to $30$, giving $$1,3,5,7,9,11,13,15,17,19,21,23,25,27,29.$$ From that list, we now remove the multiples of $3$: $$1,5,7,11,13,17,19,23,25,29.$$ And finally, we take away the multiples of $5$: $$1,7,11,13,17,19,23,29.$$ This remaining list gives all the integers $m$ between $1$ and $30$ for which $gcd(m,30)=1$. There are eight integers on this list, so $\varphi (30)=8$.

   5. Here, we can use the same approach that worked for calculating $\varphi (27)$, which was also a power of a prime number. Knowing that $p$ is the only prime factor of $p^k$, if we have an integer $m$ between $1$ and $p^k$, we know that $gcd(m,p^k)\ne 1$ exactly when $m$ is a multiple of $p$.

      There are $p^k$ integers from $1$ to $p^k$, and exactly $\frac{p^k}{p}=p^{k-1}$ of them are multiples of $p$, namely all integers of the form $np$, where $n$ ranges from $1$ up to $p^{k-1}$. Subtracting off all of these multiples of $p$ from the total, we get $p^k-p^{k-1}$ integers $m$ between $1$ and $p^k$ not divisible by $p$. This tells us $$\varphi (p^k)=p^k-p^{k-1}.$$ Notice that this agrees with the $k=1$ case worked out in Example 5, as well as the case $p=3,k=3$ that we solved in part (c).

5. Modular arithmetic is good for much more than RSA. If an equation is true in the integers, we can “reduce it mod $n$" to get a congruence mod $n$ that is also true. This is more useful in the other direction: if we start with an equation and prove it has no solutions mod $n$ for some choice of $n$, then there are no integer solutions. This helps because we can exhaustively try every possibility for the variables mod $n$ – there are only $n$ different choices for each variable that would give a distinct result.

   Prove that each of the following equations has no integer solution by showing each one has no solutions mod $n$ for a small choice of $n$ (in each case, you can take $n\le 5$.)

   1. $6x+21y=2$

   2. $5y=x^2+2$

   3. $x^2+y^2=31$

   Solution:

   1. Here, we take the equation and reduce it modulo $3$. This gives us $$6x+21y\equiv 2(\mathrm{mod}3).$$ Now, notice that both $6$ and $21$ are congruent to $0$ mod $3$. This means $6x\equiv 0(\mathrm{mod}3)$ and $21y\equiv 0(\mathrm{mod}3)$. The congruence above then simplifies to $$0+0\equiv 2(\mathrm{mod}3).$$ In turn, this says $0\equiv 2(\mathrm{mod}3)$. By definition, this means $3\mid (0-2)$. This is impossible, as $-2$ is not divisible by $3$. This shows that the congruence has no solution modulo $3$, which means the original equation has no solution either.

   2. For this one, we take the equation and reduce it modulo $5$. Since $5y\equiv 0(\mathrm{mod}5)$ no matter what value we take for the integer $y$, we find that the congruence $5y\equiv x^2+2(\mathrm{mod}5)$ reduces to $$0\equiv x^2+2(\mathrm{mod}5).$$ Now, there are only five “different" integers modulo $5$, namely $0,1,2,3,4$. Let’s try them all as possible values of $x$ to see what $x^2+2$ comes out to: $$\begin{array}{rl}{0}^2+2& \equiv 2(\mathrm{mod}5)\\ 1^2+2& \equiv 3(\mathrm{mod}5)\\ 2^2+2& \equiv 6\equiv 1(\mathrm{mod}5)\\ 3^2+2& \equiv (-2{)}^2+2\equiv 6\equiv 1(\mathrm{mod}5)\\ 4^2+2& \equiv (-1{)}^2+2\equiv 3(\mathrm{mod}5).\end{array}$$ We just tried all possible values for $x$ modulo $5$, and none of them gave us $x^2+2\equiv 0(\mathrm{mod}5)$. Therefore, the congruence has no solutions, which implies the equation $5y=x^2+2$ has no integer solutions.

   3. For this equation, reducing modulo $4$ works. When we do this, we end up with $$x^2+y^2\equiv 31\equiv 3(\mathrm{mod}4).$$

      Since $0,1,2,3$ represent all the different possible values modulo $4$, we can try each of these as values for $x$ to see what the possibilities for $x^2$ will be: $$\begin{array}{rl}{0}^2& \equiv 0(\mathrm{mod}4)\\ 1^2& \equiv 1(\mathrm{mod}4)\\ 2^2& \equiv 4\equiv 0(\mathrm{mod}4)\\ 3^2& \equiv 9\equiv 1(\mathrm{mod}4).\end{array}$$ Therefore, no matter what $x$ is, we know that $x^2\equiv 0(\mathrm{mod}4)$ or $x^2\equiv 1(\mathrm{mod}4)$. Similarly, these are the only two possible values for $y^2$ modulo $4$. Putting that together, the possibilities for $x^2+y^2$ modulo $4$ will be: $$\begin{array}{rl}0+0& \equiv 0(\mathrm{mod}4)\\ 0+1& \equiv 1(\mathrm{mod}4)\\ 1+0& \equiv 1(\mathrm{mod}4)\\ 1+1& \equiv 2(\mathrm{mod}4)\end{array}$$ None of these possibilities gives us an answer of $3$ modulo $4$, so $x^2+y^2\equiv 3(\mathrm{mod}4)$ has no solutions. In turn, this implies $x^2+y^2=31$ has no integer solutions either.

6. In RSA, Nick is not supposed to reveal $\varphi (N)$, because knowing it allows everybody to calculate Nick’s private key. It’s time to put on our cryptanalysis hats and try to break the system. If we know $p$ and $q$, the two factors of $N$, it becomes easy to calculate $\varphi (N)$. However, as we know, finding those factors is really hard. Maybe there’s an easier way to calculate $\varphi (N)$? We’re going to see the answer is no.

   1. Suppose you have a magic machine that takes $N$ and instantly calculates $\varphi (N)$. How can you use the values of $N$ and $\varphi (N)$ to calculate $p+q$?

   2. Once you know both $p+q$ and $N=pq$, how can you calculate both $p$ and $q$? (Hint: Consider the quadratic formula applied to the polynomial $x^2-(p+q)x+pq$.)

   What this means is that an efficient way for calculating $\varphi (N)$ leads to an efficient way of factoring $N$, and everyone believes that factorization is really hard.

   Solution:

   1. If we have the values of $N=pq$ and $\varphi (N)=(p-1)(q-1)=pq-p-q+1$, the value of $p+q$ is easy to obtain. Indeed, notice that $N-\varphi (N)=pq-(pq-p-q+1)=(p+q)-1$.

      So, if we know $N$ and a machine has given us the value of $\varphi (N)$, then we can put $N-\varphi (N)+1$ into a calculator and find the value of $p+q$.

   2. As suggested in the hint, let’s see what the quadratic formula says about the roots of the polynomial $x^2-(p+q)x+pq$. By that formula, the roots are $$\begin{array}{rl}x& =\frac{-(-(p+q))\pm \sqrt{(-(p+q){)}^2-4pq}}{2}\\ & =\frac{(p+q)\pm \sqrt{p^2+2pq+q^2-4pq}}{2}\\ & =\frac{(p+q)\pm \sqrt{p^2-2pq+q^2}}{2}\\ & =\frac{(p+q)\pm \sqrt{(p-q{)}^2}}{2}\\ & =\frac{(p+q)\pm |p-q|}{2}.\end{array}$$ The absolute value makes things a little complicated, but because of the $\pm$ in the quadratic formula, in every case the two roots come out to be $p$ and $q$. Therefore, if you use the known numbers $N=pq$ and $p+q$, applying the quadratic formula to $x^2-(p+q)x+pq$ gives the values of $p$ and $q$ as the answers. This may be the single coolest application of the quadratic formula ever to exist!

7. Like any cryptosystem, the security of RSA can be compromised if used incorrectly. Suppose Nick and Bahaa have the same value of $N$, but different encryption exponents $e_1$ and $e_2$.

   1. Explain how Nick and Bahaa can work out each other’s private keys, and therefore read each other’s encrypted messages.

   2. Suppose Shefaza comes along and sends the same message $M$ to both Nick and Bahaa (maybe it’s Shefaza’s credit card number). Suppose also that $gcd(e_1,e_2)=1$. Now, let’s say Diana comes along and reads the ciphertexts that Shefaza sent to Nick and Bahaa, say $C_1\equiv M^{e_1}(\mathrm{mod}N)$ and $C_2\equiv M^{e_2}(\mathrm{mod}N)$.

      Explain how Diana can use this information to recover Shefaza’s plaintext $M$.

      Hint: Running the Extended Euclidean algorithm on $e_1$ and $e_2$ allows Diana to find integers $x$ and $y$ such that $e_{1}x+e_{2}y=1$. How can this information help her recover $M$?

   Solution:

   1. Because Nick and Bahaa have both chosen the same value of $N$, both of them started by choosing the same primes $p$ and $q$ to create $N$. In particular, both Nick and Bahaa know the value of $\varphi (N)$. Because of this, if Nick used exponent $e_1$ and Bahaa used exponent $e_2$, Nick could look up $e_2$ and find Bahaa’s decryption exponent $d_2$ by running the Extended Euclidean algorithm on $\varphi (N)$ and $e_2$. In the same way, Bahaa can calculate Nick’s decryption exponent $d_1$.

   2. As mentioned in the hint for the question, since $gcd(e_1,e_2)=1$, Diana can use the known values of $e_1$ and $e_2$ to find integers $x$ and $y$ such that $e_{1}x+e_{2}y=1$, by running the Extended Euclidean algorithm.

      At this point, Diana should take the ciphertexts $C_1$ and $C_2$ that she intercepted and compute $C_1^x\cdot C_2^y(\mathrm{mod}N)$. Here is what happens when she does this: $$C_1^x\cdot C_2^y\equiv (M^{e_1}{)}^x\cdot (M^{e_2}{)}^y\equiv M^{e_{1}x}\cdot M^{e_{2}y}\equiv M^{e_{1}x+e_{2}y}\equiv M^1(\mathrm{mod}N).$$

      So, if Diana calculates the unique integer $M$ between $0$ and $N-1$ such that $M\equiv C_1^x\cdot C_2^y(\mathrm{mod}N)$, then $M$ will be the desired plaintext.

      Remark: There’s a minor point to mention here: we only talked about what it means to raise numbers to positive exponents mod $N$. Here, $x$ or $y$ could be negative. There’s a way to make sense of negative exponents mod $N$, which you can investigate if you are curious! But don’t worry: the argument given here can indeed be made valid.