## Math 310: Problem set 10

Instructions: This problem set is due Tuesday, November 28, 2006. Your goal is not only to give correct answers but to communicate your ideas well. Make sure you use good English.
1. The number 708714834137 is the product pq of two primes p and q, with phi(pq)=(p-1)(q-1)=708713126820.
• Given pq and (p-1)(q-1), show how to find p+q, and determine p+q if pq and (p-1)(q-1) are as given above.
• Answer: Since pq-(p-1)(q-1)+1=p+q, we see p+q = 708714834137 - 708713126820 + 1 = 1707318.
• Given pq and p+q, show how to find p and q, and thereby find a factorization of 708714834137.
• Answer: Since (p-q)2 = (p+q)2-4pq = 1707318^2 - 4*708714834137 = 80075416576, taking the square root we have p-q = 282976, hence p = ((p+q)+(p-q))/2 = 995147 and q = ((p+q)-(p-q))/2 = 712171. Thus 708714834137 factors as 995147*712171. [Alternatively, if pq=a and p+q=b, then q=a/p so b=p+a/p or bp=p2+a, hence p2-bp+a=0, which can be solved using the quadratic formula.]

Note: The identity (p-q)2 = (p+q)2-4pq allows you to solve any quadratic equation. If the solutions of x2-cx+d=0 are r and s, then x2-cx+d = (x-r)(x-s) = x2-(r+s)x+rs, so c=r+s and d=rs. From the identity we get (r-s)2 = (r+s)2-4rs = c2-4d, so r-s = sqrt(c2-4d), so r = ((r-s) + (r+s))/2 = (c + sqrt(c2-4d))/2 and s = ((r+s) - (r-s))/2 = (c - sqrt(c2-4d))/2.

2. Consider the cubic equation x3-3ax-b=0 in the case that a=2 and b=6. Cardano's solution is to use the identity (u+v)3-3uv(u+v)-(u3+v3) = 0. If you can find u and v such that uv = a and u3+v3 = b, then x=u+v is a solution to the cubic.
• Show that from the two equations
```uv = a
u3+v3 = b
```
you can obtain a single equation (u3)2+c(u3)+d=0. What values do you get for the coefficients c and d?
• Answer: v=a/u so u3+(a/u)3 = b, hence u6+a3 = bu3, or (u3)2-b(u3)+a3=0. Thus c = -b = -6 and d = a3 = 8.
• Solve (u3)2+c(u3)+d=0 for u3, then get u by taking cube roots, and get v from uv = a. What solution x=u+v does this give you? Verify your solution by plugging it into x3-6x-6. What value do you get? (Due to round-off error, it shouldn't be zero, but it should be close.)
• Answer: The solutions to (u3)2-6(u3)+8=0 are u3 = 2 and u3 = 4, so u is either 21/3 (in which case v = a/u = 2/21/3 = 22/3 = 41/3), or u is 41/3 (in which case v = a/u = 2/41/3 = 2/22/3 = 21/3). Either way, x = u+v = 21/3 + 41/3. Plugging this on my calculator into x3-6x-6 gives -5.6810-11, which is pretty close to 0.
3. Cardano's solution for x3-3ax-b=0 using the identity (u+v)3-3uv(u+v)-(u3+v3) = 0 can be used to solve any cubic. Consider the cubic x3+cx2+dx+e=0. If you substitute x=y-c/3 in for x and simplify, you get an equation of the form y3+fy+g=0 in which the squared term has been eliminated, and which you can thus solve using Cardano's method. If y = n is a solution to y3+fy+g=0, then from x = y - c/3 we get that x = n - c/3 is a solution to the original equation, x3+cx2+dx+e=0.
• In the case of x3+3x2-3x-11=0, what should you substitute in for x, as described above, to eliminate the x2 term?
• Answer: Substitute x = y - 3/3 = y - 1.
• After making your substitution, you should have a polynomial in y with no y2 term. What is your polynomial? [Hint: It should look familiar!]
• Answer: We get x3+3x2-3x-11 = (y - 1)3 + 3(y - 1)2 - 3(y - 1) - 11 = (y3 - 3y2 + 3y - 1) + 3(y2 - 2y + 1) - 3(y - 1) - 11, which simplifies to y3 - 6y - 6.
• Give a solution to x3+3x2-3x-11=0.
• Answer: If we substitute y-1 in for x, from x3+3x2-3x-11=0 we get y3 - 6y - 6 = 0, which from the previous problem has the solution y = 21/3 + 41/3. Thus x = (21/3 + 41/3) - 1 is a solution for x3+3x2-3x-11=0.
4. Here is a message encrypted using the RSA method with a modulus of m = 10 and an encryption exponent of e = 3. Find the decryption exponent d (explain and show how you find it), and decipher the message: 020116168500800201141019030988091403. (Hint: Letters were converted into strings of digits by converting A to 01, B to 02, etc. However, I'm not telling you the block size. What is the biggest possible block size if m = 10? What block size must have been used?)
5. Answer: Since m=10 and blocks must give distinct elements of Z/mZ, we can't take blocks giving numbers bigger than 9. Thus the block size must be 1. Since phi(10) = phi(2)phi(5) = 1*4 = 4, and since de == 1 (mod phi(m)), we find that d = 3. Thus the string decrypts as follows:
```ciphertext 020116168500800201141019030988091403
plaintext  080116162500200801141019070922091407, giving numeral encoded letters
08 01 16 16 25 00 20 08 01 14 10 19 07 09 22 09 14 07 or
H  A  P  P  Y     T  H  A  N  J  S  G  I  V  I  N  G
```
(The J instead of a K is a typo on my part...)