Download Raji 4.4: 1. Find the six smallest perfect numbers. This is the same

Raji 4.4:
1. Find the six smallest perfect numbers.
This is the same as …nding the six smallest Mersenne primes. So we must
check Mp = 2p 1 for primality for p = 2; 3; 5; 7; 11; 13; : : :. Hopefully we
won’t have to go too far out!
M2 = 22 1 = 3 is prime, so the smallest perfect number is 21 22
3
M3 = 2
5
M5 = 2
496.
2
1 = 7 is prime, so the second perfect number is 2
1 = 6.
3
1 = 28.
4
5
2
1 = 31 is prime, so the third perfect number is 2
2
1 =
M7 = 27 1 = 127 is prime (it is not divisible by 2, 3, 5, 7, or 11), so the
fourth perfect number is 26 27 1 = 8128.
M11 = 211 1 = 2047. Is this prime? According to Theorem 51, any
divisor must be of the form 22m + 1. So try m = 1, that is, 23. Indeed,
2047 = 23 89, so M11 is not a prime.
M13 = 213 1 = 8191. Is this prime? According to Theorem 51, any
divisor mustpbe of the form 26m + 1. We only have to check divisors that
are at most 8191 = 90:5 so we need only check 27, 53, and 79. We only
have to check for prime divisors, so we don’t have to check 27 because it
is divisible by 3. We compute 8191=53 = 154:55 and 8191=79 = 103:68, so
M13 is prime and the …fth perfect number is 212 213 1 = 33 550 336.
It’s also true that any prime divisor of a Mersenne number is congruent
to 1 modulo 8. So we didn’t have to check 53.
M17 = 217 1 = 131 071. Is this prime? Theorem 51 says that the divisors
are of the form 34m + 1. Theorem 51 says to check 35, 79, 103, 137, 171,
205, 239, 273, 307, 341. This is far enough because 3752 = 140 625. We
needn’t check 35, 171, 205, or 273 because 35 and 205 are divisible by 5,
and 171 and 273 are divisible by 3. So we check 79, 103, 137, 239, 307,
and 341. Actually, 341 is divisible by 11, so we don’t have to check it.
Moreover, 307 3 (mod 8), so we don’t have to check it either.
131071=79 = 1659:1,
131071=103 = 1272:5,
131071=137 = 956:72,
131071=239 = 548:41,
131071=307 = 426:94,
131071=341 = 384:37.
So M17 is prime and the sixth perfect number is 216 217
1
1 = 8; 589; 869; 056.
2. Find the eighth perfect number.
Is he serious? I guess you could look it up. It turns out that M19 and
M31 are prime but M23 and M29 are not. So the eighth perfect number is
230 231 1 = 2; 305; 843; 008; 139; 952; 128
3. Find a factor of 21001
1.
If 1001 were a prime, this would be a di¢ cult problem. If we can …nd a
factor of 1001, then it’s an easy problem. In fact, although 1001 is not
divisible by 2, 3, or 5, it is divisible by 7. In fact, 1001 = 7 11 13. Isn’t
that amazing? That’s why 1001 is a magic number, as in the book “One
thousand and one nights”. So 27 1 = 127 is a factor of 21001 1. See
the proof of Theorem 50 in Raji.
4. We say n is abundant if (n) > 2n. Prove that if n = 2m 1 (2m 1)
where m is a positive integer and 2m 1 is composite, then n abundant.
We have (n) =
2m 1 (2m 1) because
is multiplicative. But
m 1
m
2
=2
1 and if 2m 1 has a factor other than 1 and 2m 1,
then (2m 1) > 2m 1 + 1 = 2m , so (n) > (2m 1) 2m = 2n.
5. Show that there are in…nitely many even abundant numbers.
From the previous exercise, it su¢ ces to show that 2m 1 is composite
for in…nitely many m. But 2m 1 is divisible by 22 1 = 3 for every even
positive integer m.
Actually, every multiple of 12 is abundant. Indeed, any multiple of an
abundant number is abundant. That’s because (ab)
(a) b, which
follows easily from the simple fact that if a0 is a factor of a, then a0 b is a
factor of ab.
6. Show that there are in…nitely many odd abundant numbers.
Can you …nd any odd abundant numbers? I believe the smallest one is
945 = 33 5 7. Indeed, (945) =
33 (5) (7) = 40 6 8 = 1920. Now
we can use the observation in the previous exercise that any multiple of
an abundant number is abundant. So, in particular, any odd multiple of
945 is an odd abundant number.
7. Determine whether M11 is prime.
Didn’t we already do that in Exercise 1? There we showed that M11 =
23 89.
8. Determine whether M29 is prime.
We failed to do that in Exercise 2. Since M29 = 229 1 = 536 870 911
is not prime, maybe it’s easy to …nd a factor. Any factor must be of the
form 58m + 1. So divide by 59, 117, 175, 233, . . . . We can skip 117 and
175 because they are not prime.
536 870 911=59 = 9099506:966.
2
536 870 911=233 = 2304167
so M29 is divisible by 233, hence not prime.
n
9. Find all primes of the form 22 + 5 where n is a nonnegative integer.
0
1
2
Well, 22 +5 = 7 is a prime, and 22 +5 = 9 is not a prime, and 22 +5 = 21
is not a prime. So maybe 7 is the only one. Are the rest divisible by 3?
3
How about 22 + 5 = 261? Yes. Look modulo 3. If m is even, then 2m
is a power of 4, hence congruent to 1 modulo 3. As 5 is congruent to 2
n
modulo 3, we have 22 + 5 is congruent to 1 + 2 = 3 modulo 3, hence is
divisible by 3, if n > 0. Yay!
3