Find all positive integers k such that there exists positive integers a,b such that a2+4=(k2−4)b2.
Problem
Source: 2023 Singapore MO Round 2 Senior Q2 / Junior Q5
Tags: number theory, Diophantine equation
24.06.2023 14:24
Let x=a+kb2 ofcurse x is natural then this can be written as x2−kxb+b2+1 Wlog max(x,b)=x then: If x=b then we esily have k=3 and x=b=1 Now let (x0,b0) be the smaller sollution then there is a second sollution (x1,b0) with: x1+x0=kb0 and x1x0=b2+1 from this two we get x1 natural. Also sinse (x0,b0) is the smaller sollution then x1>=x0 but we have: b_0^2+1=x_1*x_0\geqslant x_0^2\geqslant (b_0+1)^2>b_0^2+1 contradiction
21.11.2023 07:29
just look to case a divisible by 2 and not ,and use if a^2+b^2 divisible by p=4k+3 where p is prime \rightarrow a divisble by p and b
19.08.2024 15:47
P2nisic wrote: Let x=\frac{a+kb}{2} ofcurse x is natural then this can be written as x^2-kxb+b^2+1 Wlog max (x,b)=x then: If x=b then we esily have k=3 and x=b=1 Now let (x_0,b_0) be the smaller sollution then there is a second sollution (x_1,b_0) with: x_1+x_0=kb_0 and x_1x_0=b^2+1 from this two we get x_1 natural. Also sinse (x_0,b_0) is the smaller sollution then x_1>=x_0 but we have: b_0^2+1=x_1*x_0\geqslant x_0^2\geqslant (b_0+1)^2>b_0^2+1 contradiction Doesn't seem to quite work as written but does lead to the famous Vieta jumping finish once the nice sub is made.