# Proof for Finding Amicable Numbers

4.3 Proof for Finding Amicable Numbers

It’s important to define some terms before we begin the discussion on the proof for finding amicable numbers. Definition: Let oo(n) be the sum of the proper divisors of the positive integer n. for instance,

So,

go (16) = 1+2+3+4+8 = 18.

n is a perfect number if oo(n) = n (such as n = = 6, since oo(6) =

6).

n is an abundant number if oo(n) > n (such as n = 16, since do(16) > 16).

n is a deficient number if oo(n) < n (such as n = 8, since go(8) < 8).

Ibn Qurra also lists 10 important propositions for constructing his proof. For brevity’s sake, we list the 5th, 6th, and 9th propositions which he stipulates:

Proposition 5 If p is prime and a = 2″p, then go(a) = 2n+1 12″pp. Further, if p = 2n+11, then a is perfect, if p < 2n+¹ − 1, then a is abundant, and if p > 2n+¹ – 1, then a is deficient 34

Proposition 6 Let a = 2″ pip2. Let P1 P2 be different prime numbers and not 2. Let k = 2n+¹1+ (2n+1 1) (p₁ + p2). Then, oo(a) = k + 2″ P1P2 – P1P2. Also, if p₁p2 <k, a is abundant, and if pip2 > k, then a is deficient. Proposition 9 If a, b, c, d are all natural numbers with corresponding ratio 1: 2: 4 8,

then

d(a + d – 1) = c[(d(a + d) − 1) − (d+c− 1)(b + c− 1)]

(propositions 7 and 8 are used to prove this).

We follow the proof outlined in [2]:

3 This is a generalization of Euclid’s Elements IX, 3.

4The English translation of Ibn Qurra’s original Proposition 5 is as follows: “When one adds a sequence of numbers following each other in double progression starting from the unit, and when one gets a certain sum, then when one multiplies the greatest numbers added by a prime number other than two, the number obtained by this multiplication will be a perfect number, if the prime number is equal to the sum obtained [from the sequence]. But, if the prime number is smaller than this sum, the product will be an abundant number; and if the prime number is larger than the sum, the product will be a deficient number. And, the magnitude of excess, if it is an abundant number – or if its defect, if it is a deficient number – is equal to the difference between the sum and the prime number mentioned above” [21].

2n+1 -1, p2 2n+1 -1, P3 2n+1(2n+1+ Theorem 4.1. Let n> 1 be an integer. Let pi 2n-2)-1 be three prime numbers (but not 2). Then, a₁ = 2″p₁p2 and a2 = 2″p3 are amicable numbers, i.e o(a₁) = a2 and o(a₂) = a₁. – =

Proof. We want to show that a2 − o(a₂) = A2 go(a₁) a₁ = a2 – a₁, which implies that Proposition 5, we have that a₁, which implies that o(a2) = a₁, and σo(a₁) =a2. Let us first find a2 – o(a₂). By –

– σo (a₂) =a₂ − P3 + (2n+¹ − 1) ⇒ a₂ – oo(a₂) = P3 − (2n+¹ − 1) = 2n+1 2n+1 [2n+1 +2n-2] − 1 − (2¹+¹ − 1) = 2n+¹[(2n+¹ − 1 + 2²−2)]. = So, a2 σ0 (α₂) 2n+¹[(2n+¹ − 1) + 2n-2]. We find 00(a₁) conclude that go(a₁) – a₁ also equals 2n+¹[(2n+¹ − 1) + 2¹−2]. Now, let us find a2 a₁. Firstly, note that Thabit ibn Qurra rewrites p2 as 2n-1 + 2n a₁ in a similar fashion and – 1.

Thus,

a₂ – a₁ = 2¹ (P3 – P1P2) = 2″ [(2n+¹(2n+¹ + 2n−2) − 1) − (2n+¹ +2″ − 1) (2¹−1 + 2″ − 1)] = 2n+¹ [(2n-2 + 2n+¹) — 1]. –

The first equality holds by the definitions of a₁ and a2. The second equality holds by the definitions of P₁, P2, and p3. The last equality holds by Proposition 9, taking a = 2n-2 b = 2n-1, c = 2″, and d 2n+1. So, by Proposition 6, we get that o(a₁) = = a2 and o(a₂) = a1.

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