Source code in the 8086 programming language

LIMIT:	equ	20000		; Maximum value
	cpu	8086
	org	100h
section	.text
	mov	ax,final	; Set DS and ES to point just beyond the
	mov	cl,4		; program. We're just going to assume MS-DOS
	shr	ax,cl		; gave us enough memory. (Generally the case,
	inc	ax		; a .COM gets a 64K segment and we need ~40K.)
	mov	cx,cs
	add	ax,cx
	mov	ds,ax
	mov	es,ax
calc:	mov	ax,1		; Calculate proper divisors for 2..20000
	mov	di,4		; Initially, set each entry to 1. 
	mov	cx,LIMIT-1	; 2 to 20000 inclusive = 19999 entries 
	rep	stosw 
	mov	ax,2		; AX = outer loop counter
	mov	cl,2
	mov	dx,LIMIT*2	; Keep inner loop limit ready in DX
	mov	bp,LIMIT/2	; And outer loop limit in BP
.outer:	mov	bx,ax		; BX = inner loop counter (multiplied by two)
	shl	bx,cl		; Each entry is 2 bytes wide
.inner:	add	[bx],ax		; divsum[BX/2] += AX
	add	bx,ax		; Advance to next entry
	add	bx,ax		; Twice, because each entry is 2 bytes wide
	cmp 	bx,dx		; Are we there yet?
	jbe	.inner 		; If not, keep going
	inc	ax
	cmp	ax,bp		; Is the outer loop done yet?
	jbe	.outer		; If not, keep going
show:	mov	dx,LIMIT	; Keep limit ready in DX 
	mov	ax,2		; AX = outer loop counter
	mov	si,4		; SI = address for outer loop
.outer:	mov	cx,ax		; CX = inner loop counter
	inc	cx
	mov	di,cx		; DI = address for inner loop
	shl	di,1
	mov	bx,[si]		; Preload divsum[AX]
.inner: cmp	cx,bx		; CX == divsum[AX]?
	jne	.next		; If not, the pair is not amicable
	cmp	ax,[di]		; AX == divsum[CX]?
	jne	.next		; If not, the pair is not amicable
	push	ax		; Keep the registers
	push 	bx
	push	cx
	push 	dx
	push	cx		; And CX twice because we need to print it
	call	prax		; Print the first number
	pop	ax
	call	prax		; And the second number
	mov	dx,nl		; And a newline
	call	pstr
	pop	dx		; Restore the registers
	pop 	cx
	pop 	bx
	pop	ax
.next:	inc	di		; Increment inner loop variable and address
	inc	di		; Address twice because each entry has 2 bytes 
	inc	cx
	cmp	cx,dx		; Are we done yet?
	jbe	.inner		; If not, keep going
	inc	si		; Increment outer loop variable and address
	inc	si		; Address twice because each entry has 2 bytes
	inc	ax 
	cmp	ax,dx		; Are we done yet?
	jbe	.outer		; If not, keep going.
	ret 
	;;;	Print the number in AX. Destroys AX, BX, CX, DX.
prax:	mov	cx,10		; Divisor
	mov	bx,nbuf		; Buffer pointer
.digit:	xor 	dx,dx
	div	cx		; Divide by 10 and extract digit
	add	dl,'0'		; Add ASCII 0 to digit
	dec	bx
	mov	[cs:bx],dl 	; Store in string
	test	ax,ax		; Any more?
	jnz	.digit		; If so, keep going
	mov	dx,bx		; If not, print the result
	;;;	Print string from CS.
pstr:	push	ds		; Save DS
	mov	ax,cs		; Set DS to CS
	mov	ds,ax
	mov	ah,9		; Print string using MS-DOS
	int	21h
	pop	ds		; Restore DS
	ret
	db	'*****'
nbuf:	db	' $'
nl:	db	13,10,'$'
final:	equ	$