Suppose that $ a,b$ are positive integers for which $ A=\frac{a+1}{b}+\frac{b}{a}$ is an integer.Prove that $ A=3$.
Problem
Source: Romanian TST for BMO 1991
Tags: quadratics, number theory proposed, number theory
xpmath
14.07.2009 16:43
We have $ A = \frac {a^2 + b^2 + a}{ab}$. Suppose $ A\ne{3}$ and consider the solution $ (a, b)$ with minimal $ a + b$.
Rearranging to form a quadratic in $ a$, we have
\[ a^2 + a(1 - Ab) + b^2 = 0\]
Let $ x$ be the other root of this quadratic. Then, $ x = Ab - 1 - a$, implying that $ x$ is an integer. Additionally, $ x = \frac {b^2}{a}$, implying $ x$ is positive. Thus, $ x$ is a positive integer so that $ (x, b)$ is a solution to this equation. We must have $ x\ge{a}$ so $ \frac {b^2}{a}\ge{a}$, meaning $ b\ge{a}$.
Now rearrange to form a quadratic in $ b$:
\[ b^2 - b(Aa) + a^2 + a = 0\]
Let the other root of the quadratic be $ y$. Then by the same reasoning as before, $ (a, y)$ is a solution to the equation. We must have $ y = \frac {a^2 + a}{b}$. As $ y\ge{b}$, we have $ a^2 + a\ge{b^2}$. However, $ (a + 1)^2 = a^2 + 2a + 1 > a^2 + a$, so $ b^2 < {(a + 1)^2}$, implying $ b < a + 1$.
Thus, $ a\le{b} < a + 1$, so $ a = b$. Substituting, $ A = \frac {2a^2 + a}{a^2} = 2 + \frac {1}{a}$. To be an integer, we must have $ a = 1$, which gives $ A = 3$, a contradiction.
Hence, $ A = 3$.
v235711
16.07.2009 03:58
$ A.a.b=a^2+a+b^2$. Therefore $ a|b$.Hence we must have that $ \dfrac{a+1}{b}$ in an integer.But $ a|b$.Therefore $ a=1$ and the result follows.
xpmath
16.07.2009 04:03
Sorry if I'm missing the obvious, but how do you know $ a|b$...?
pythag011
16.07.2009 04:07
v235711 wrote: $ A.a.b = a^2 + a + b^2$. Therefore $ a|b$.Hence we must have that $ \dfrac{a + 1}{b}$ in an integer.But $ a|b$.Therefore $ a = 1$ and the result follows. a=4 b=2 is a counterexample to a divides b
v235711
16.07.2009 04:27
pythag011 wrote: v235711 wrote: $ A.a.b = a^2 + a + b^2$. Therefore $ a|b$.Hence we must have that $ \dfrac{a + 1}{b}$ in an integer.But $ a|b$.Therefore $ a = 1$ and the result follows. a=4 b=2 is a counterexample to a divides b Sorry about my dumb mistake.
Rust
16.07.2009 04:40
Let $ d=(a,b),a_1=a/b,c=b/d$, then $ (a_1,c)=1$ and $ A=\frac{d(a_1^2+c^2)+a_1}{a_1c d}.$ Therefore $ d|a_1$ and $ a_1|dc^2\to a_1|d$. It mean $ a_1=d,a=d^2,b=dc,(c,d)=1, A=\frac{d^2+1+c^2}{cd}>2$. These question consider in these forum at least 2 times, that only $ A=3$ give solutions for Pell like equation $ d^2+c^2-Acd=-1$, and had infinetely many solutions.