Determine all pairs $(a,b)$ of natural numbers such that the number $$\frac{a^2(b-a)}{b+a}$$is the square of a prime number.
Problem
Source: Mathematics Regional Olympiad of Mexico West 2019 P3
Tags: number theory, Perfect Square
iniffur
09.09.2022 17:37
My try
$a^2\left(\frac{b-a}{b+a}\right)=p^2$
Let $a=pd\Longrightarrow p^2d^2\left(\frac{b-pd}{b+pd}\right)=p^2$
$\Longrightarrow d^2\left(\frac{b-pd}{b+pd}\right)=1\Longrightarrow d^2b-pd^3=b+pd$
$\Longrightarrow b=p\left(\frac{d^3+d}{d^2-1}\right)\in N$
But $~~d^2-1\nmid d^3+d\Longrightarrow d^2-1\mid p$
This only occurs if $~~~d=2, p=3\Longrightarrow a=6, b=10~~~$ (Only solution with this approach)
Pal702004
10.09.2022 13:46
#2, $a<b$ cannot be a solution. Solutions: If $p=t^2+1$ then a=$t(t^2+1),b=t(t^2-1)$ If $p=(t^2+1)/2$ then a=$t(t^2+1)/2,b=t(t^2-1)/2$ Examples: $(a,b,p)=(10,6,5);(15,12,5);(65,60,13);(68,60,17);(222,210,37);(369,360,41)\cdots$
Pal702004
10.09.2022 14:05
Sorry, my mistake, my solutions are for $\dfrac{a^2(a-b)}{(a+b)}$
Wakkis1729
10.09.2022 15:13
Pal702004 wrote: Sorry, my mistake, my solutions are for $\dfrac{a^2(a-b)}{(a+b)}$ yh, For this case, there is a solution for all primes of the form $\frac{k^2+1}2$ or $k^2+1$
TheRealRuirui
10.09.2022 15:46
Wakkis1729 wrote: Since there are infinitely many primes of the form $\frac{k^2+1}2$ and infintely many primes of the form $k^2+1$ How can you prove this?