Number Theory Problem No.1

Number Theory Level 4

Given that $\alpha \beta \gamma = 6$ , find the value of $abc$ for $a, b, c, d$ are positive integers and coprime satisfy these conditions:

$b^{a} - a^{b} = b - a = \alpha$

$b^{b} - c^{a} = c - b = \beta$

$a^{d} - c^{b} = c - a = \gamma$

Bonus: Can you find the values of $a, b, c, d$ ? Give your answer in the explanation.

Hope everyone stay safe at home!

66 70 42 30 105 110

1 solution

Pop Wong
Aug 1, 2020

a,b,c,d are positive integers and coprime
$\alpha\beta\gamma = 6$

As a, b, c, d are integers, $\alpha,\beta,\gamma$ are integers but not necessary all positive too.

$\gamma = c - a = (c - b) + (b-a) = \beta + \alpha \\ \therefore \alpha\beta\gamma = \alpha\beta(\alpha+\beta) = 6 \\ \Rightarrow \alpha = 1, \beta=2, \gamma =3 \\ \Rightarrow (a, b, c) = (a, a+1, a+3) \\ \because a,b,c,d$ are coprime $\therefore b=a+1$ and $c=a+3$ are odd and $a$ is even

Then we can check as guess your solution is $(a, b, c, d) = (2, 3, 5, 7)$

Number Theory Problem No.1

1 solution

0 pending reports