Sorry, can't remember who posted this in SAX today. [Edit: It was Nik - thanks Kerm!] I can answer it, but will leave it for others pro tem:

An infinite area is partly black, partly white. There may be a pattern, or a random coloring, you don't know.
Prove that there must be [at least] two points, exactly 5 inches apart, which are the same color.

Yes I posted it... For an additional challenge, if you solved the previous one:

You have an infinite space (3-dimensional) and every spot in it is either black, grey or white. Now you need to prove that there are at least two points which have the exactly same color and are exactly 5 inches apart. As in the last puzzle, this is true again.

For the solutions... I haven't any thoughts but it would be unfair for the ones who are still solving if someone posted the solution already...
I think I will consider the solution public in 24h and post it here/confirm someone else's solution.
Have fun!

A bit easy
Let's say you have a random black point somewhere. If the thesis is NOT true, then all the points at the circle with a radius of 5 inches must be white otherwise the thesis is true. But within the white circle, you can draw many lines of 5 inches between two white points on that circle, so the thesis must be true. I hope this prove is acceptable

The proof is fine, and is exactly the same as my first thought. Nik has messaged me a proof that is arguably even simpler:
For any equilateral triangle, with 5 inch sides, the three vertices can't be three different colors (there are only 2 colors), so at least two of them are the same color, and are of course 5 inches apart.

I agree it was easy!

I am sorry for that, I thought this was more or less a hard task. But maybe someone has another puzzle?

A regular n-sided polygon is inscribed in [fits exactly inside] a circle of radius 1. Choose one vertex [corner] of the polygon, and consider its distance from each of the other vertices [corners].
Prove that the product [multiplication result] of all these n-1 distances is exactly n.

Consider the points to be the nth roots of unity on the unit circle, and by symmetry the one point to be 1. Then the distances to the points are the magnitudes of the roots of (x-1)^n-1=0. Expand the nth power: 0=x^n+nx^(n-1)+...+nx+1-1

0=x(x^(n-1)+nx^(n-2)+...+n)

The x at the start represents 1+0i, so divide that out. The last term is n; by Vieta's formulas, the product of the distances is n.

That's pretty much it: I don't know of a proof without use of complex numbers, but there must be one!
It should perhaps be emphasised that the product of the roots of the polynomial equation is n * (-1)^(n-1). But the product of the magnitudes = the magnitude of the product, for complex numbers, so it is indeed n.

This is too complicated for me... Maybe because I haven't learned that yet, or because it is too late or because my english is not that good or I am simply not smart enough... But can't you always square an imaginary number to make it non imaginary?

Yes, the square of a purely imaginary number is always real (and negative). But the method used here by lirtosiast was based on complex numbers, which are part-real and part-imaginary. It does rely on quite a bit of knowledge of complex number properties.
I think your English is excellent, but I didn't know why you asked your question.

I thought one can do (5+2i)^2 and have 25+(-4)=21 to escape the need of using complex numbers, but I just tested it, that seems wrong.

In general (5+2x)² = 25+20x+4x², so (5+2i)²=25+20i-4 = 21+20i.
The complex numbers in the puzzle above are used because they are a 2-dimensional system (real and imaginary) whereas the ordinary real numbers are one-dimensional (the number line). So they are providing a convenient numerical codification of the two-dimensional polygon, and lirtosiast's proof is the same way I did it myself. But I was wondering whether anyone found a purely geometrical/trigonometrical proof.

I was looking for one and I had the mind that it might have to do with their average... But I can't calculate that accurately.

I'm pretty sure I could construct one given the fundamental duality between complex analysis and trig, but it doesn't seem particularly fun or interesting.

Relevant facts are:

• The chord-length function for a chord of angle t is 2sin(t/2).
  
• the angle through m adjacent sides of a regular n-gon is 2mπ/n.
  
• Special case: the side lengths of a regular n-gon inscribed in a unit polygon are 2*sin(π/n)
  

Solve for integers m & n the equation 3^m = 2^n + 1.
Remember that, for a mathematician, to solve an equation means to find all the solutions, with a demonstration that they are a complete solution set.

Not sure if this is a good answer, but it seems that m and n can only be 1...
Anything larger, and 3 would grow exponentially with a higher base, so faster than the 2, so the equation wouldn't be true for any bigger numbers, and same goes with negative numbers (it would shrink exponentially)... as for 0, well because of that +1, that one wouldn't work either...

(m,n)=(1,1) and (3,2) are solutions; all numbers less than that can be checked manually. (Nonpositive integers cannot give solutions, because nonpositive powers of 2 and 3 all lie in (0,1], which means there's no room for them to be 1 apart.) Now I will show that these are the only solutions.

Assume that m>3 and n>2. Note that n must be odd: 2^n = 3^m - 1 => 2^n = -1 mod 3, but only odd powers of 2 are -1 mod 3. m must be even, because only even powers of 3 are +1 mod 4.

Then the equation becomes a Pell equation of form x^2 - 2*y^2 = 1, with x^2 = 3^m and 2*y^2 = 2^n, whose solutions are Pell numbers. Since n>3, y must be a perfect power of 2.

The Wikipedia page for Pell numbers says that the only perfect Pell number powers are 0, 1, and 169; therefore no Pell number is a perfect power of two, giving no other solutions.

I'm afraid I'd have to say that doesn't read like a proof. The 3 would grow faster than the 2 is not a relevant comment, since they are not raised to the same power. This has caused you to miss one of the simple solutions: see lirtosiast's post.

Yeah, I guess I assumed they had to be the same, oops

Hmmm... <Teacher> Not too sure if I can award full marks for a proof which quotes "Wikipedia says"! </Teacher>
I did learn some Pell equations in my degree course, but can't remember too much about them after nearly 40 years! But it isn't necessary to go to that level.
A) If just one of m,n is negative, then one of 3^m, 2^n is an integer and the other isn't, so the equation is false.
B) If both m&n are negative, then 3^m and 2^n are both between 0&1, so the equation is false.
So we know we are looking for non-negative integers [You had this part covered]
C) If m is odd, 3^m = -1 (mod 4), so 2^n = 2 (mod 4), so n=1 as no other powers of 2 have this property. Therefore m=1, n=1 is the only solution for odd m. [You had this nailed down too].
D) If m is even, say m=2r, where r is a non-negative integer. Therefore 2^n = 3^2r - 1 = (3^r - 1)(3^r + 1). Those two brackets are both integers, and they are factors of a power of 2. Therefore they are themselves powers of 2, by Unique Factorization. But they differ by 2. The only two powers of 2 which differ by 2 are 2 & 4, so 3^r=3, r=1 and m=2. So the only solution for even m is m=2, n=3.
You were correct there are no other solutions, but I hope you find this to be a more complete proof.
Many thanks for your post which was an interesting take on the problem.