Find the number of pairs of positive integers $a$ and $b$ such that $a\leq 100\,000$, $b\leq 100\,000$, and $$ \frac{a^3-b}{a^3+b}=\frac{b^2-a^2}{b^2+a^2}. $$
Problem
Source: 2020 Caucasus Mathematical Olympiad
Tags: algebra
Math-wiz
16.03.2020 10:52
Haha, componendo and dividendo
This is equivalent to $a^5=b^3$ after simplification.
Let $a^5=b^3=k$
Then, $k$ is a $lcm(3,5)=15th$ power of an integer.
$b^3=c^{15}$ for some $c\in\mathbb{N}$
$b=c^5\leq 10000$
$c\leq 10$
So the only solutions are
$(1,1),(2^3,2^5),(3^3,3^5)\ldots (10^3,10^5)$
Dr Sonnhard Graubner
16.03.2020 16:03
Hello, you will get the equation $$a^5-b^3=0$$Sonnhard.
dikhendzab
20.03.2020 12:39
After multiplication of the equation you will get: $a^3b^2-a^5+b^3-ba^2 = a^3b^2+a^5-b^3-ba^2$ We will have now $2(a^5-b^3)=0$ or simple $a^5-b^3=0$ or $a^5=b^3$. The trivial solution is $a=b=1$. $a, b$ are less or equal than $100 000$, so the solutions are: $(a, b) = (1, 1), (2^3, 2^5).... (9^3, 9^5), (10^3, 10^5).$ That is all.
celilcelil
26.06.2020 14:06
So easy. (a, b)=(n^3, n^5) where 0 <n <11