Month: April 2023

He (Ibn Sad) said: Muhammad IbnUmar al-Aslami related to us; he
said: Fulayh Ibn Sulayman related to me on the authority of Sa’id Ibn
Ubayd Ibn al-Sabbàq, he on the authority of Abu Sa'id al-Khudri; he said: After the arrival of the Prophet, may Allah bless him, in al-Madinah, when a person was on the verge of death, we went to him (Prophet) and informed him. He came and prayed for his salvation. When the man expired, he (Prophet) returned with those who were with him; some times he stayed till the corpse was interred. Some times the Prophet, may Allah bless him, was thus detained for long. When we feared that it would be troublesome for him, some people said to others: By Allah ! we should not inform him before a person's death . (After this) we informed him only after the death (of a person) so that he was spared this trouble of being detained. He (al-Khudri) said: We did accordingly, and informed him only after a person had died. Then he would come, say funeral prayers, and pray for his salvation. Some times he left and at other times he stayed till the corpse was interred. For a time it was our practice. Then the people said: By Allah ! it would have been better if we had not informed the Apostle of Allah, may Allah bless him, and had taken the bier to his house and sent him a message and he would have said the funeral prayers near his house. It would have been more convenient and easier for him. He (al-Khudri) said: We did accordingly. Muhammad IbnUmar said: This place came to be known as the
place of biers because the dead bodies were taken there. Then it became a
custom among the people to carry the biers there and to say the funeral
prayers at that place. This is the practice till today.

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.