Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$a\times b = 8 \times (a + b)$

I have used Wolfram Alpha and it has given me 14 integer solutions.

But how can we find those solutions ? Which method ?

edit: integer solutions are as following:

$a = [-56, -24, -8, 0, 4, 6, 7, 9, 10, 12, 16, 24, 40, 72]$

$ b = [7, 6, 4, 0, -8, -24, -56, 72, 40, 24, 16, 12, 10, 9]$

share|improve this question

2 Answers 2

up vote 6 down vote accepted

Hint: $$ab-8(a+b)=0 \iff (a-8)(b-8)=64$$

share|improve this answer
    
the amount of integers can be found this way but finding the exact values of those integers is still bugging me. –  Selim Arikan Feb 18 '13 at 13:56
    
@SelimArikan Use factorization of 64. For example, $$(a-8)(b-8)=32\cdot 2$$ So one proposed solution of this equation is $a-8=32$, $b-8=2$, so $a=40, b=10$. –  tetori Feb 19 '13 at 1:51
    
thank you for your explanation –  Selim Arikan Feb 19 '13 at 7:25

So, $$a=\frac{8b}{b-8}=\frac{8(b-8)+64}{b-8}=8+\frac{64}{b-8}\iff (b-8)\mid64$$

Now, $64=2^6$ has $6+1=7$ positive divisors

So, we have $7$ solutions if we consider the only the positive values of $b-8$.

So, if we include all the non-zero integers, we shall have $2\cdot7=14$ divisors.

Clearly, $b-8\ne0$ else $a$ will be infinite.

So, the values of $b-8$ are $\pm 2^i$ for integer $i\ge0$ and $\le6$

Hence,

if $i=0,b-8=\pm1,b=9,7;$

if $i=1,b-8=\pm2,b=10,6;$

if $i=2,b-8=\pm4,b=12,4;$

if $i=3,b-8=\pm8,b=16,0;$

if $i=4,b-8=\pm16,b=24,-20;$

if $i=5,b-8=\pm32,b=40,-24;$

if $i=6,b-8=\pm64,b=72,-56;$

share|improve this answer
    
But there are only three negative solutions for a, [-56, -24, -8] and how to find those values ? –  Selim Arikan Feb 18 '13 at 12:02
    
@SelimArikan, observe that we have $7$ negative and $7$ positive values for $b-8$. Sorry for the delay. –  lab bhattacharjee Feb 18 '13 at 15:00
    
thank you for your explanation –  Selim Arikan Feb 19 '13 at 7:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.