(The following problem is open in the sense that the answer to part (b) is not currently known.) Let $n$ be a positive integer that is not a perfect square. Find all pairs $(a,b)$ of positive integers for which there exists a positive real number $r$, such that $$r^a+\sqrt{n} \ \ \text{and} \ \ r^b+\sqrt{n}$$are both rational numbers. Let $n$ be a positive integer that is not a perfect square. Find all pairs $(a,b)$ of positive integers for which there exists a real number $r$, such that $$r^a+\sqrt{n} \ \ \text{and} \ \ r^b+\sqrt{n}$$are both rational numbers.
Problem
Source: Simon Marais Mathematics Competition 2023 Paper B Problem 4
Tags: number theory
17.10.2023 16:34
Here is a solution of part (a): We claim that the only solutions are $\boxed{a=b}$. Clearly these work, so now WLOG assume $a>b$. Firstly, by replacing $r$ by $r^d$ if necessary, we can assume $\gcd(a,b)=1$. Also, $r \neq 1$ since $\sqrt{n}$ is not rational. Claim 1: $r>1$. This is the part that we missed in the actual exam :-( Proof: Assume FTSOC that $r<1$. Let $r^a+\sqrt{n}=q$ and $r^b+\sqrt{n}=s$ be (positive) rationals; then $q<s$ but $(q-\sqrt{n})^b=(s-\sqrt{n})^a$. Taking conjugates in $\mathbb{Q}(\sqrt n)$, we get $(q+\sqrt n)^b=(s+\sqrt n)^a$, which is absurd because $a>b$ and $s+\sqrt n > q+\sqrt n >1$. $\blacksquare$ Claim 2: $r \in \mathbb{Q}(\sqrt n)$. Proof: Follows from Bezout because $r^a,r^b \in \mathbb{Q}(\sqrt n)$ and $\gcd(a,b)=1$. $\blacksquare$ Claim 3: $r=c-d \sqrt n$ where $c,d$ are positive rational numbers. Proof: Write $r=c-d \sqrt n$ where $c,d$ are some rationals. One of $a,b$ must be odd, so assume $a$ is odd (the other case works similarly). Setting the $\sqrt n$ part of $(c-d \sqrt n)^a +\sqrt n$ to $0$, we get $$\sum_{i=0}^{\frac{a-1}{2}} \binom{a}{2i+1} c^{a-2i-1}d^{2i+1}n^i = 1$$Since each $a-2i-1$ is even, $c^{a-2i-1} \geq 0$, and $d^{2i+1}$ has the same sign as $d$, so the above can hold only if $d>0$. Now since $r>0$, we also get $c>0$. $\blacksquare$ Now, if we directly expand $r^{a-b}=(c-d \sqrt n)^{a-b}$, we get that it is equal to $c_0-d_0 \sqrt n$ for some positive rationals $c_0,d_0$ (this is because all the irrational terms would have an odd exponent of $-d \sqrt n$, and rational terms would have even exponent). Now $r>1$ $\implies$ $r^{a-b}>1$ $\implies$ $c_0-d_0 \sqrt n>1$ $\implies$ $c_0 >1$. Recall that $r^a=q-\sqrt n$ and $r^b=s -\sqrt n$. Then we have $$q-s=r^a-r^b=r^b(r^{a-b}-1)=(s- \sqrt n)(c_0-1 - d_0 \sqrt n) = s(c_0-1)+d_0 n -((c_0-1)+sd_0) \sqrt n$$which is a contradiction since the LHS is rational while the RHS has a non-zero $\sqrt n$ part (because $(c_0-1)+sd_0>0$). Hence there are no solutions when $a \neq b$. $\blacksquare$