For which values of $\theta$ does this equation $x^{\cos\theta} +y^{\sin\theta }=1$ have solutions in integers?

For which values of $\theta$ does this equation $$x^{\cos\theta} +y^{\sin\theta}=1$$ have solutions in integers ?

Note : $x, y$ integers, $\theta$ is real number.

• if $\theta =0$ then $x=0$ and $y\not =0$ arbitrary.. – Empty May 28 '15 at 16:19
• if $\theta =\pi/2$ then $y=0$ and $x\not =0$ arbitrary.. – Empty May 28 '15 at 16:20
• What's the source of this question? – punctured dusk May 28 '15 at 16:21
• Why do you say solutions in integers, then say $x,y,\theta$ are real? – Ross Millikan May 28 '15 at 16:22
• @SmailHamidani. If you want geometrical interpretation, check if here. – user164524 May 28 '15 at 16:32

If either of the exponents is strictly positive $(\alpha > 0)$, then $(0,1)$ or $(1,0)$ is a solution, because $0^{\alpha} + 1^{\beta} = 1$ for any $\beta$. This occurs for all $\theta \in (-\pi/2, \pi)$, and leaves two cases:

• One exponent is zero and the other is strictly negative: $\theta\in\{-\pi/2,\pi\}$. In this case there is no solution... any integer raised to the zeroth power is $1$ (if defined), but no integer raised to a negative power is $0$.
• Both exponents are strictly negative: $\theta \in (\pi, 3\pi/2)$.

In the second case, we want $$\frac{1}{x^\alpha} + \frac{1}{y^\sqrt{1-\alpha^2}}=1$$ for some $\alpha \in (0,1)$ and integers $x,y$. If $x$ or $y$ is $0$ or negative, the expression is undefined, and if $x$ or $y$ is $1$, then the left-hand side is strictly greater than $1$; so we must have $x,y\ge 2$. Fix $x$ and $y$ and consider the behavior of the left-hand side as we vary $\alpha$: $$f(\alpha;x,y)=e^{-\alpha \log x}+e^{-\sqrt{1-\alpha^2}\log y},$$ so $$f'(\alpha;x,y)=-\log x e^{-\alpha\log x}+\frac{\alpha \log y}{\sqrt{1-\alpha^2}}e^{-\sqrt{1-\alpha^2}\log y},$$ which goes from $-\log x < 0$ at $\alpha=0$ to $+\infty > 0$ at $\alpha=1$; and $$f''(\alpha;x,y)=(\log x)^2 e^{-\alpha\log x}+\left(\frac{\alpha^2 (\log y)^2}{1-\alpha^2}+\frac{\log y}{\sqrt{1-\alpha^2}}-\frac{\alpha^2 \log y}{(1-\alpha^2)^{3/2}}\right)e^{-\sqrt{1-\alpha^2}\log y},$$ which is always positive. (*I think.) So for each $x,y$, there is a unique minimum of $f(\alpha;x,y)$ with respect to $\alpha$, at which $$\frac{\alpha\log y}{y^{\sqrt{1-\alpha^2}}}=\frac{\sqrt{1-\alpha^2}\log x}{x^{\alpha}}.$$ Depending on whether the value of $f$ at this point is less than, equal to, or greater than $1$, there will be $2$, $1$, or $0$ solutions in $\alpha$ to the original equation for this $(x,y)$ pair. Checking this out numerically, we find that there are no solutions for $(x,y)=(2,2)$ and $(x,y)=(2,3)$, and there are two solutions for each other $(x,y)$ pair with $x,y\ge 2$.

• Thank you for your answer, since we must to check to the numerical analysis it's seems that it's a big problem and fine equation. – zeraoulia rafik May 28 '15 at 17:36
• @ mjqxxxx , does this equation related to last fermat theorem ? – zeraoulia rafik May 28 '15 at 21:38
• can I take this as a finally and enough answer or should be waiting others answers ? – zeraoulia rafik May 28 '15 at 23:07

For all couple of natural numbers (n,m) we have $\frac{1}{n} + \frac{1}{m}$$\leq n^{cos\theta}+ m^{sin\theta}$$\leq n+m$ because
$\frac{1}{n}$$\leq n^{cos\theta}$$\leq{n}$ and the same for the sinus; therefore, when $\frac{1}{n} + \frac{1}{m}$<1 one has by continuity, a value of $\theta$ (actually infinitely many by periodicity) for which the asked equality is verified. Hence if $3\leq n, m$ there are always solution for (n,m). What values of $\theta$? I don’t know. For (2,2) and (2,3) there are no solution because the minimum of $2^{cos\theta}$+$2^{sin\theta}$ and $2^{cos\theta}$ +$3^{sin\theta}$ are both greater than 1. However for (2,m) with $4\leq{m}$ we have solutions (it was good and my edition was bad so I erased it)

• Gomez : thank you for your answer , But the main problem is the values of $\theta$ not the values of $cos\theta$ or $sin\theta$ – zeraoulia rafik May 28 '15 at 23:29
• I don't see, Smail, how to do that without a calculator. – Piquito May 28 '15 at 23:47
• do you meant that your answer good ? – zeraoulia rafik May 28 '15 at 23:54
• I mean, Zeraoulia, that for each of infinitely many couples (n,m) you have an infinity of solutions, with a positive minimum value $\theta_{0}$ and the other ones differing of the first by multiples of the periodes. And you have infinitely many of these $\theta_{0}$ (besides, as solution of a trascendental equation so you can't find out the solutions; at least I believe you can't...., I don't discard anyway, I am wrong about it) – Piquito May 29 '15 at 15:31
• so edit your answer mayeb it will be good – zeraoulia rafik May 29 '15 at 19:40