Suppose that a and b are positive integers such that $a-b=6$ and $\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9$. Find the smallest possible value of b.

pinklemonade Oct 2, 2019

#2**+1 **

\(a-b=6\)

\(\text{gcd}\left(\frac{a^3+b^3}{a+b}, ab\right) = 9\)

\(\frac{(a+b)(a+b)^2}{(a+b)}\)

\((a+b)^2\)

So now it is \(\text{gcd}\left((a+b)^2, ab\right) = 9\)

So now we can plug in a - b = 6

a = 6 + b

\(\text{gcd}((2b+6)^2,b^2+6b)\)

Ok notice how b^2 + 6b must be divisible by 9, so we just keep guess and checking until we hit our first integer that is divisible by nine for b.

1 = 7

2 = 16

3 = 27

So b=3 works for b^2 + 6b,

now we test for (2b+6)^2

We get 144, which is divisible by 9.

So the answer should be **3.**

My solution is very bogus, it may be wrong, or it may be right! But most likely wrong. Can someone check?

.CalculatorUser Oct 2, 2019

#3**+1 **

a=1; b=1;c= gcd((a^3 + b^3)/(a + b), a * b); if(c==9, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;

a = 9 and b = 3

Guest Oct 3, 2019

#4**0 **

Nice job guest! Cool how you can easily program something that can solve problems like this.

CalculatorUser
Oct 3, 2019