# Solutions to Trigonometric Problem Class

Is there a way to prove that the general solution of:

$\sin^2 \pi x + \sin^2 \frac {ab\pi}{x} = 0$

is:

$x = \pm 1,\pm a, \pm b, \pm ab$

and more specifically to derive the proof analytically, without relying on appeals to the logical behavior of the function? (i.e. "Well I know the zeros of $\sin \pi x$ are all the integers, so the answers must be integers")

As an example, I know that the zeros of:

$\sin^2 \pi x + \sin^2 \frac {6\pi}{x} = 0$

are

$\pm 1, \pm2,\pm3$, and $\pm6$ because the only time $\frac 6x$ is an integer and $x$ is an integer is when $x=\pm 1, \pm2,\pm3$, or $\pm6$, but that is unsatisfying and far from a rigorous proof.

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The statement is true only if $a$ and $b$ are primes; otherwise $ab$ can have more divisors than the ones you listed. –  Peter Košinár Jul 19 '13 at 15:48
fair. Technically those four pairs are zeros for any integers, they're only an exhaustive list for primes. –  Jason Nichols Jul 19 '13 at 15:49
Now, if you disregard the knowledge of "the only zeroes of $\sin \pi x$ are integers" as being non-analytic; what properties of $\sin$ do you allow to be used in the proof? –  Peter Košinár Jul 19 '13 at 15:51
For the case $ab=6$, seems plenty rigorous enough. –  André Nicolas Jul 19 '13 at 15:55
Peter, any a priori knowledge about the outcome merely gives proof by circular reasoning. "I know the zeros are integers, so the zeros must be integers." It still doesn't give a method for finding the zeros if they are unknown. –  Jason Nichols Jul 19 '13 at 15:59

Since both terms are squared, the sum is zero iff each term is 0 (this is just a norm property, if you are trying to be as analytical as possible). So we can take each term and write out when it is 0, and just take the intersection of those two sets. The first term is 0 at the integers only (as a proof, use the fact that sin is periodic), and the second is 0 when $ab/x$ is an integer. So the only solutions are integer divisors of $ab$.

Are you also looking for a proof that the only zeros of $\sin \pi x$ are the integers, or is the above sufficient?

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Still looking for a solution that doesn't involve independently testing against all the elements of the sets. Basically because it relies on the assumption that there will be a match and then verifying there's a match. This doesn't prove the solution holds true for all $ab$ that are integers or even primes. –  Jason Nichols Jul 19 '13 at 18:02

$\sin^2\pi x+\sin^2 \frac{ab\pi}{x}=0 \Rightarrow \sin^2x=0$ and $\sin^2 \frac{ab\pi}{x}=0$ since $\sin^2(\cdot)$ is always non-negative because otherwise, if $\sin^2(\pi x)>0$, then $\sin^2 \frac{ab\pi}{x}<0$ and vice-versa, which is absurd for $x\in \mathbb{R}$. This implies $x\in \mathbb{N}$ and $x|ab$. Hence the solutions are $x\in\{{}_{-}^+d\in \mathbb{Z}:\ d|ab \}$.

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More formally stated, but still relies on guess and test factorization. Still gets an upvote. :) –  Jason Nichols Jul 19 '13 at 18:03
Actually I'm not guessing, see the nice thing that the individual $\sin^2(\cdot)$s become $0$ follow from the non-negativity of $\sin^2(\cdot)$. The rest is just the consequence. –  Samrat Mukhopadhyay Jul 19 '13 at 18:08
right, but it still means if you're given $ab$ as opposed to $a$ and $b$, the only way to figure it out is to guess and test, i.e. if $ab=903$ what are the zeros to the function. What I'm trying to root out is the reliance on a priori knowledge of $ab$, and truly solve for $a$ and $b$ as opposed to looking for collisions between two non trivially large sets. –  Jason Nichols Jul 19 '13 at 18:14
I see, you are trying to get the roots explicitly in terms of $a,\ b$. –  Samrat Mukhopadhyay Jul 19 '13 at 18:20
yessir. I want to prove the theorem without relying on checking all members of every possible set. I know the answer because a priori I can push any value $ab$ into the equation and know that the zeros are the factors $a$ and $b$, but nothing I've tried so far has given me a quadratic equation-like solution where the factors are calculated as opposed to checked as intersections of arbitrarily large sets. –  Jason Nichols Jul 19 '13 at 18:35