Equation Involving GCD

What I did is by completing the squares where we solve b in terms of a and their GCDs....

$b = \dfrac{a \gcd (a, b) \pm \sqrt{[a \times \ gcd(a,b)]^2 - 4 (a + \gcd(a,b)]^3))}}{2}$

We let $d = gcd(a, b)$

And we have to make the expression inside the radical integral. By completing the squares...

$(ad)^2 - 4(a+d^3) = k^2$ ( $k$ is an integer)

$\implies ad - \left( \dfrac{2}{d} \right)^2 - 4d^3 = k^2 + \left(\dfrac{4}{d^2} \right)$

If $d | 2$ , then $d = 1$ or $2$ , which implies $a = 5$ and 4 respectively.

What is not complete: For $d > 2$ , I checked the base cases for $d = 3, 4$ and what I found out and conjectured:

If $d | 2$ , then there exists a and b that are positive integers that satisfy the equation and for $d$ more than 3, there is not. When there is a solution, but it doesn't satisfy the conditions.

I'll try to prove this....

Okay, I didn't solve this problem mathematically. I wrote the python code:

1
2
3
4
5
6

from fractions import gcd

for a in range(1,100):
    for b in range(1,100):
        if a + b**2 + (gcd(a,b))**3 == a*b*gcd(a,b):
            print [a,b]

and it returns a small solution set:

[4, 2]

[4, 6]

[5, 2]

[5, 3]

4 + 5 = 9

If a and b have a common factor>1, then we let $a=p^{\alpha}.A,b=p^{\beta}.B$ and $d=p^{min{\alpha,\beta}}.D$ , where A,B and D are coprime to p and alpha and beta are >0.Then by looking at $v_p{LHS}$ and $v_p{RHS}$ , we conclude that $\alpha=2.\beta$ is the only possible case.Consequently $a=b^2$ , yielding the solution $a=4$ .Now if a and b are coprime ,then d=1 and we factor in the following manner:

$b^2-1+2=a(b-1)$ , which means that $b-1|2$ , leaving the only other solution where a=5 and b=2 or 3.

So the answer is 9.

Set f (a, b) = a + b^2 + (gcd (a, b))^3 - a b gcd (a, b),

f (4, 2) = 4 + 2^2 + 2^3 - (4)(2)(2) = 16 - 16 = 0;

f (5, 2) = 5 + 2^2 + 1^3 - (5)(2)(1) = 10 - 10 = 0;

f (4, 6) = 4 + 6^2 + 2^3 - (4)(6)(2) = 48 - 48 = 0;

f (5, 3) = 5 + 3^2 + 1^3 - (5)(3)(1) = 15 - 15 = 0;

Minimum magnitude f (b + 1, b) = 2 always happens example when 32 x 31 of search using Excel is implemented. As told by the question, a value of b = 2 for a = 4 or 5 shall be the only answer. Therefore two distinct 'a' of 4 + 5 = 9.