0
$\begingroup$

I am asked to prove that there are no rational point solutions to $a^{2}+b^{2} = 3^{k}$ where $k$ is an odd, positive integer.

I have already proved there are no rational solutions (aside 0,0,0) for $a^{2}+b^{2}=3$

The proof for that worked by contradiction. I proved $a$ and $b$ had to be even, and that $3c^{2}\equiv3*0$ mod 4 to maintain the equality, thus c had to be even too.

So in this $a^{2}+b^{2} = 3^{k}$ I think the way to solve it is say that $3c^{2}\equiv3^{k}c^{2}$ (mod 4) for $k$ an odd, positive integer. Then the proof works just the same as the $a^{2}+b^{2}=3$ proof.

I'm struggling to write a direct proof of this equivalence. I tried using the definition $3=4m+n$ and expand $3^q=(4m+n)^{q}$ But I can't reduce that to an equation where I divide by 4 and am left with $n$.

So my question is, assuming my approach to the main question is correct, how can I write a direct proof for $3c^{2}\equiv3^{k}c^{2}$ (mod 4) for $k$ an odd, positive integer.

  • 0
    You call $a$ and $b$ 'rational' at one point, and 'even' at another. Do you mean that $a$ and $b$ are actually integers? – 2017-02-09

2 Answers 2

1

Since $k$ is an odd positive integer. $(-1)^k = -1$.

$$3^k \equiv (-1)^k \equiv -1 \equiv 3 \mod 4$$

  • 0
    Darn this is clever, thank you. I guess the reason $3^k\equiv(-1)^k$ mod 4 is that 3 and -1 are separated by 4, so you only only change the number of fours that evenly divide into either quantity, the remainder stays the same. – 2017-02-09
0

Assume you've proved there are no rational solutions to $$a^2 + b^2 = 3$$ Let $k>1$ be an odd positive integer.

If $a,b$ are rational numbers such that $$a^2 + b^2 = 3^k$$ then letting \begin{align*} x &= \frac{a}{3^{(k-1)/2}}\\[6pt] y &= \frac{b}{3^{(k-1)/2}}\\[6pt] \end{align*} you get $$x^2 + y^2 = 3$$ contradiction.

Including the known case $k=1$, it follows that:

If $k$ is an odd positive integer, there are no rational numbers $a,b$ such that $a^2 + b^2 = 3^k$.