Find all pairs $(a,b)$ of prime positive integers $a,b$ such that number $A=3a^2b+16ab^2$ equals to a square of an integer.
Problem
Source: 2020 Greek JBMO TST p3
Tags: number theory, Perfect Square
i3435
14.11.2020 23:22
What's the structure of the greek tst test? This seems offly trivial for a $3$.
This factorizes to $ab(3a+16b)$. If this is a square, $ab|3a+16b$. Thus either $a|16$ and $b|3$ or $a=b=19$. This means either $(a,b)=(2,3)$ or $(a,b)=(19,19)$
pi_quadrat_sechstel
14.11.2020 23:22
$ab(3a+16b)$ is a perfect sqare. If $a=b$, the number $19a$ is a perfect square. Since $a$ is prime $a=b=19$. If $a\neq b$, we must have $a\mid 3a+16b$ and $b\mid 3a+16b$. Since $a,b$ are different primes, we have $a\mid16$ and $b\mid3$. Thus $a=2,b=3$. Obviously, $(2,3),(19,19)$ are solutions.
parmenides51
14.11.2020 23:25
i3435 wrote: What's the structure of the greek tst test? This seems offly trivial for a $3$. this is JBMO TST problem, for the Junior Balkan MO (JBMO) , it comes right after the Junior National Math Olympiad (it has 3 rounds, the 3rd is the final)
Euler...
08.03.2024 23:28
$A=ab(3a+16b)=k^2$ there are 2 cases , if a=b then $(a;b)=(19;19)$ , second case is $a\neq b$ GCD(a;b)=1 a|3a+16 also b|3a+16b and here there is only one solution (2;3)=(a;b)