MATH 313: SOLUTIONS HW2
Problem 1
(a) 30941
We use the Miller-Rabin test to check if it prime. We know that the smallest number which is a
strong pseudoprime both base 2 and base 3 is 1373653; hence, if 30941 passes the test for both
base 2 and base 3 it will have to be prime since 30941 < 1373653.
Base 2: 30490 = 22 × 7735.
27735 ≡ 5826 (mod 30941).
215490 ≡ −1 (mod 30941) ⇒ pass the test.
Base 3:
37735 ≡ 5826 (mod 30941).
215490 ≡ −1 (mod 30941) ⇒ pass the test.
Therefore 30941 is prime and we are done!
(b) 33918
Use Miller-Rabin test to check primality.
Base 2: 33918 = 2 × 16959.
216959 ≡ 8186 (mod 33919).
233918 ≡ 20571 (mod 33919).
Fail the test! so 33919 is composite.
After checking the legendre symbols, we should find the following factor base:
{2, 3, 5, 7, 17, 19, 31, 37, 43, 47, 53}.
Also note that
√
33919 = 184 and we find that sieving between 170 and 203 is sufficient.
I omit the table with the computation of x and Q(x). I will just use a combination which works for
factoring:
Q(173)Q(177)Q(188)Q(193)Q(203)
= (−2 × 3 × 5 × 7 × 19)(−2 × 5 × 7 × 37)(3 × 52 × 19)(2 × 32 × 5 × 37)(2 × 36 × 5)
= 24 × 310 × 56 × 192 × 372 .
We can now look at the gcd to check if we get non-trivial factors:
gcd((173 × 177 × 188 × 193 × 203) − (22 × 35 × 53 × 19 × 37), 33919) = 317
gcd((173 × 177 × 188 × 193 × 203) + (22 × 35 × 53 × 19 × 37), 33919) = 107
Hence, 33919 = 107 × 317.
(c) 68569
Use Miller-Rabin test to check primality.
Base 2: 68568 = 23 × 8571.
28571 ≡ 14870 (mod 68569).
217142 ≡ 50444 (mod 68569).
234284 ≡ 1546 (mod 68569).
268568 ≡ 58770 (mod 68569).
Fail the test! so 68569 is composite.
After checking the legendre symbols, we should find the following factor base:
{2, 3, 5, 7, 17, 19, 23, 29, 31, 53, 61}.
√
Also note that 68569 = 262 and we find that sieving between 242 and 282 is sufficient.
I omit the table with the computation of x and Q(x). I will just use a combination which works for
factoring:
Q(262)Q(263)Q(264)Q(265)
= (3 × 52 )(23 × 3 × 5)(72 × 23)(23 × 32 × 23)
= 26 × 34 × 54 × 72 × 232 .
We can now look at the gcd to check if we get non-trivial factors:
gcd((262 × 263 × 264 × 265) − (23 × 32 × 52 × 7 × 23), 68569) = 191
gcd((262 × 263 × 264 × 265) − (23 × 32 × 52 × 7 × 23), 68569) = 359
Hence, 68569 = 191 × 359.
(d) 838861
Use Miller-Rabin test to check primality.
Base 2: 838860 = 23 × 8571.
2209715 ≡ 2048 (mod 838861).
2419430 ≡ −1 (mod 838861).
Pass the test! so 838861 can be a prime.
Base 3:
3209715 ≡ 456516 (mod 838861).
2419430 ≡ 231416 (mod 838861).
2838860 ≡ 478816 (mod 838861).
Fail the test! so 838861 is a composite.
After checking the legendre symbols, we should find the following factor base:
{2, 3, 5, 7, 11, 13, 17, 19, 29, 31, 37}.
Also note that
√
838861 = 916 and we find that sieving between 896 and 936 is sufficient.
I omit the table with the computation of x and Q(x). I will just use a combination which works for
factoring:
Q(917)Q(923)
= (22 × 3 × 132 )(22 × 33 × 112 )
= 24 × 34 × 112 × 132 .
We can now look at the gcd to check if we get non-trivial factors:
gcd((917 × 923) − (22 × 32 × 11 × 13), 838861) = 397
gcd((917 × 923) + (22 × 32 × 11 × 13), 838861) = 2113
Hence, 838861 = 371 × 2113.
(e) 1016801
Use Miller-Rabin test to check primality.
Base 2: 1016800 = 25 × 31775.
231775 ≡ −1 (mod 1016801).
Pass the test! so 1016800 can be a prime.
Base 3:
3209715 ≡ 509141 (mod 1016801).
363550 ≡ 294140 (mod 1016801).
3127100 ≡ 776112 (mod 1016801).
3254200 ≡ 991348 (mod 1016801).
3508400 ≡ 152972 (mod 1016801).
31016800 ≡ 791371 (mod 1016801).
Fail the test! so 1016800 is a composite.
After checking the legendre symbols, we should find the following factor base:
{2, 5, 7, 11, 19, 29, 31, 37, 41, 3, 47, 53}.
Also note that
√
1016801 = 1008 and we find that sieving between 978 and 1038 is sufficient.
I omit the table with the computation of x and Q(x). I will just use a combination which works for
factoring:
Q(991)Q(999)Q(1011)Q(1022)
= (−25 × 5 × 7 × 31)(−24 × 52 × 47)(23 × 5 × 7 × 19)(19 × 31 × 47)
= 212 × 54 × 72 × 192 × 312 × 472 .
We can now look at the gcd to check if we get non-trivial factors:
gcd((991 × 999 × 1011 × 1022) − (26 × 52 × 7 × 19 × 31 × 47), 1016801) = 4051
gcd((991 × 999 × 1011 × 1022) + (26 × 52 × 7 × 19 × 31 × 47), 1016801) = 251
Hence, 1016801 = 4051 × 251.
(f) 1149847
Use Miller-Rabin test to check primality.
Base 2: 1149846 = 2 × 574923.
2574923 ≡ 923023 (mod 1149847).
21149846 ≡ 372808 (mod 1149847).
Fail the test! so 1149847 is a composite.
After checking the legendre symbols, we should find the following factor base:
{2, 3, 13, 17, 19, 23, 41, 61, 67, 71, 73}.
Also note that
√
1149847 = 1072 and we find that sieving between 1012 and 1131 is sufficient.
I omit the table with the computation of x and Q(x). I will just use a combination which works for
factoring:
Q(1077)Q(1085)
= (2 × 712 )(2 × 34 × 132 )
= 22 × 34 × 132 × 712 .
We can now look at the gcd to check if we get non-trivial factors:
gcd((1077 × 1085) − (2 × 32 × 13 × 71), 1149847) = 521
gcd((1077 × 1085) + (2 × 32 × 13 × 71), 1149847) = 2207
Hence, 1149847 = 521 × 2207.
Problem 2
Here are 6 possible factors:
(1) Q(103) × Q(107) = 210 × 72 .
2 potential factors:
gcd(103 × 107 + 210 × 72 , 10553)
gcd(103 × 107 − 210 × 72 , 10553)
(2) Q(103) × Q(110) × Q(111) = 26 × 72 × 132 × 172 .
2 potential factors:
gcd(103 × 110 × 111 + 26 × 72 × 132 × 172 , 10553)
gcd(103 × 110 × 111 − 26 × 72 × 132 × 172 , 10553)
(3) Q(107) × Q(110) × Q(111) = 210 × 72 × 132 × 172 .
2 potential factors:
gcd(107 × 110 × 111 + 210 × 72 × 132 × 172 , 10553)
gcd(107 × 110 × 111 − 210 × 72 × 132 × 172 , 10553)
Problem 3
Note that the form p2k − 2047qk2 = ±Qk−1 ⇔ p2k ≡ ±Qk−1 (mod 2047). In order to use the quadratic sieve, we want the product of some (±Qk−1 ) to be a perfect square. Observe that in the
Qk+1 given, we already have a perfect square when pk = 181 and Qk+1 = 32 . Therefore, we can
use this one to try to factorize 2047.
Following the observation above, we have:
1812 ≡ 32 (mod 2047).
Now, we find that gcd(181 + 3, 2047) = 23 and 2047/23 = 89. Hence, we get 2047 = 23 × 89.
The period is at least 5.
Problem 4
Let’s look at the first few convergents of
√
154421.
Note: You were supposed to compute the convergents using the algorithm used in class. I will omit
this step in the solution, but you should be able to obtain the same 6 first convergents by going
through the algorithm. [Steps of the algorithm: (1) compute αi ’s and ai ’s, (2) find pi ’s and qi ’s.
You now have the convergents and are able to compute Qi ’s.]
First convergent: 392 ⇒ 3922 − 154421 × 12 = −757. Not a perfect square.
Second convergent: 393 ⇒ 3932 − 154421 × 12 = 28 = 22 × 7. Not a perfect square and cannot
get a perfect square with a product of previous convergents.
Third convergent: 11003/28 ⇒ 110032 − 154421 × 282 = −55 = 5 × 11. Not a perfect square
and cannot get a perfect square with a product of previous convergents.
Fourth convergent: 154435/393 ⇒ 1544352 − 154421 × 3932 = 196 = 22 × 72 = 142 . Perfect square. We can compute the gcd to find out if it gives us a factorization: gdc(154435 +
14, 154421) = 1 and gdc(154435 − 14, 154421) = 154421. So it doesn’t give us a factorization!
We continue looking at convergents...
Fifth convergent: 474308/1207 ⇒ 4743082 − 154421 × 12072 = −565 = −5 × 113. Not a perfect
square and cannot get a perfect square with a product of previous convergents.
Sixth convergent: 628743/1600 ⇒ 6287432 − 154421 × 16002 = 49 = 72 . Perfect square. We
can compute the gcd to find out if it gives us a factorization: gdc(628743 + 7, 154421) = 503 and
154421/503 = 307. Hence we get 154421 = 307 × 503.