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# number theory: 7 conjectures

## Question

All the numbers below should be assumed to be positive integers.
Definition. An abundant number is an integer n whose divisors add up to more than
In.
Definition. A perfect number is an integer n whose divisors add up to exactly In.
Definition. A deficient number is an integer n whose divisors add up to less than In.
Example. 12 is an abundant number, because 1 + 2 + 3+ 4 + 6+12 = 28 and 28 >
2x12. However, 14 is a deficient number, because 1 + 2 + 7 + 14 = 24, and 24 <
2 x 14.
Your task is to consider the following conjectures and determine, with proofs,
whether they are true or false.
Conjecture 1. A number is abundant if and only if it is a multiple of 6.
Conjecture 2. If n is perfect, then kn is abundant for any k in N.
Conjecture 3. If p1 and p2 are primes, then p1/p2 is abundant.
Conjecture 4. If n is deficient, then every divisor of n is deficient.
Conjecture 5. If n and m are abundant, then n + m is abundant.
Conjecture 6. If n and m are abundant, then nm is abundant.
Conjecture 7. If n is abundant, then n is not of the form pm for some natural m and
prime p.
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The definitions should be in comparison of 2n not in.

All the numbers below should be assumed to be positive integers.

Definition. An abundant number is an integer n whose divisors add up to more than

In.

Definition. A perfect number is an integer n whose divisors add up to exactly In.

Definition. A deficient number is an integer n whose divisors add up to less than In.

Example. 12 is an abundant number, because 1 + 2 + 3+ 4 + 6+12 = 28 and 28 >

2x12. However, 14 is a deficient number, because 1 + 2 + 7 + 14 = 24, and 24 <

2 x 14.

Your task is to consider the following conjectures and determine, with proofs,

whether they are true or false.

Conjecture 1. A number is abundant if and only if it is a multiple of 6.

Conjecture 2. If n is perfect, then kn is abundant for any k in N.

Conjecture 3. If p1 and p2 are primes, then p1/p2 is abundant.

Conjecture 4. If n is deficient, then every divisor of n is deficient.

Conjecture 5. If n and m are abundant, then n + m is abundant.

Conjecture 6. If n and m are abundant, then nm is abundant.

Conjecture 7. If n is abundant, then n is not of the form pm for some natural m and

prime p.

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If p1 and p2 are primes then p1/p2 cannot be an integer and therefore not abundant (Definition 1).

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i think he ment multiplication not division there.

1) false. 28000 :: 1 +2 +4 +5 +7 +10 +14 +16 +25 +28 +32 +125 +224

875 +1000+1120+1750+2000+2800 +4000 +5600 +7000 +14000 = 40638, definitely abundant.

2) true, though i have no idea how to prove it at this point

3) true, with the exception of 2 and 3.

multiplication of two primes will produce a larger result than addition.

4) true, follows from 2 assuming that can be proven.

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1. False, as proven by phil1882. A smaller example is 40 < 1+2+4+5+8+10+20.
2. True for k>1. For each divisor di of n, consider the larger number kdi which is a divisor of kn. Sum(kdi) = k*Sum(di) = 2kn, but 1 is also a divisor of kn which has not yet been accounted for. So the sum of kn's divisors is at least 2kn+1, which makes kn abundant.
3. Not sure what you mean. Should it be p1*p2? In that case, 2*3 = 6 is perfect, all others are deficient. If p1 = p2, the divisors of p1*p2 are 1, p1, and p12. The sum 1+p1+p12 < p1+p1+p12 = 2*p1+p12 </= p1*p1+p12 = 2*p12 = 2*p1*p2, so p1*p2 is deficient. If p1 < p2, the divisors of p1*p2 are 1, p1, p2, and p1*p2. The sum 1+p1+p2+p1*p2 </= p2+p2+p1*p2 = 2p2+p1*p2 </= p1*p2+p1*p2 = 2*p1*p2, with equality iff p1 = 2 and p2 = 3.
4. True. The negation is equivalent to "there is a non-deficient positive integer n and a positive integer k such that kn is deficient", which is easily disproven using the method in my proof for conjecture 2.
5. False. 40 and 12 are abundant, but 40+12 = 52 > 1+2+4+13+26 is deficient.
6. True. If n is abundant, then nm is abundant for any m.
7. False. Every n>1, abundant or otherwise, has a prime factor p, and therefore is of the form pm for some natural m.

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nicely done for 2.for seven i'm not sure if thats what hes asking.

i think he means that if n is abundant, then it must not have one prime raised to some power.

that is if n is abundant, it must be something like 2^4*3^2*5^3....

but i agree based on 2 its definitely false.

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