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#1




Creating a square with a "nonhypothenus area", under certain restriction
There's a tool for math visualization and experimentation called a Geoboard. You use rubber bands on a board with an evenly space grid of pegs to create geometric figures.
One game on this for younger math students is to create squares with various areas and realise that you can make any square with an area that is a hypothenuse number. (Exactly how is left as an exercise for the reader.) But you can also just use the bands to create lines from one peg to another, and if you do two such lines in parallel and combine with orthogonal lines with the same spacing, you also get a square, which can of course be one of the ones with an "hypothenuse area", but also one with a noninteger area. What I'd like to know is if it's possible to create a whole number nonhypothenuse area this way, such as 3, but the challenge is so far defeating me. 
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#2




An illustration, if I managed to do things right:
https://goo.gl/photos/RSQwDsroqyZTVXix5 
#3




First: by a "hypotenuse number", I assume you mean any number of the form a^{2} + b^{2}. If that's correct, then I think you can show that you can't form squares of an arbitrary side length, and in particular, 3 is impossible. But I don't know whether this generalizes to all nonhypotentuse numbers.
Consider the two lines that form the opposite sides of this square. We can assume that one of them passes through (0,0), while the second line has an equation of ax + by + c = 0. (The first line will obey a similar equation, but with c = 0.) Since the second line connects two points with integer coordinates, it can be shown that we can put this equation in a form where a, b, and c are all relatively prime integers. ("Relatively prime" means that they have no common divisor other than 1.) We can then use known results from plane geometry to find that the distance between these two lines is D = c/√(a^{2} + b^{2}). If these two lines are to form opposite sides of a square of area 3, then the distance between them must be D = √3, which implies that 3 (a^{2} + b^{2}) = c^{2}. This implies that c is a multiple of 3, or c = 3d for some d; which means that a^{2} + b^{2} = 3 d^{2}. Thus, a^{2} + b^{2} is also a multiple of 3. But square numbers are always either multiples of 3 or 1 more than multiples of 3. (In fancy language, they're either 0 or 1 mod 3.) The only way for a^{2} + b^{2} to be a multiple of 3 is for a and b to both be divisible by 3. But we said above that a, b, and c had no common divisors greater than 1. This is a contradiction, and so we can conclude that it's impossible to form a square of area 3 in this way. This proof relies critically on the fact that a^{2} + b^{2} can't be a multiple of D^{2} = 3 without a and b also both being multiples of D^{2} = 3. If D was some other nonhypotenuse number, this proof technique might not be possible. It still holds for D^{2} = 7 (the next nonhypotenuse number), but I don't want to work out D^{2}= 11 (the following one) before I've had my coffee. More later, if I have further thoughts. ETA: oh, and please don't hesitate to ask for clarification if any of the above is unclear; I've used a bunch of jargon and glossed over a few results, but I'm happy to fill those gaps in if need be. Last edited by MikeS; 05312017 at 06:56 AM. 
#4




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So numbers that can't be written as a^{2} + b^{2} was what I was after. And I enjoyed confirming for myself the correctness of the bits you glossed over. 


#5




I'm pretty sure that the general proof is just the same as the general proof that the square root of any integer other than a perfect square is irrational.

#6




I think its possible to prove you can't make non square and nonhypotenusal area squares in the peg board geometrically.
To construct any square, the rule is that each side extended must pass over two holes. The dX and dY between two holes MUST be integer. So the two holes must have the angles of a hypotenusal triangle.. Any right angle triangle constructed from that base line and the grid, must have the angles being the hypotenusal angle, which then must be reflected in the orientation of the desired square.. So by similar triangles argument, the square must be orientated the same as a hypotenusual square. This contradicts the ability to make it at a nonhypotenusual angle and therefore you cannot make the square differently to the set of square squards and hypotenusual squares you already had. Last edited by Isilder; 05312017 at 10:56 AM. 
#7




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#8




The proof is pretty straightforward for odd numbers. Any odd number of the form x^{2} + y^{2} must be congruent to 1 mod 4. This means that any "nonhypothenuse" number must be 3 mod 4. It is a basic number theory exercise to show the square of any integer is either 0 mod 4 (even) or 1 mod 4 (odd). Thus no prefect square is 3 mod 4
So no perfect square is 3 mod 4 and all nonhypothenuse numbers are 3 mod 4 therefore there is no square with integral sides that has a nonhypothenuse area. 
#9




Just to clarify. x^{2} + y^{2} is congruent to 1 mod 4 is an if and only if statement for odd numbers.



#10




As for even numbers, a number is "nonhypothenuse" if and only if it is in the form 2^{m}u where u is an odd number congruent to 3 mod 4.
Assume the side of a square of length 2^{n}v; v is odd. The area will be 2^{2m}w where w is an odd number congruent to 1 mod 4. Since this is not 3 mod 4 then no square of evenlengthed side can have a nonhypothenuse area. 
#11




Unless I have missed something, it seems simple. You can simply introduce coordinates so that the pegs are all at distance 1. Assume one edge of square goes from (a,b) to (c,d). Then the length of that edge is sqrt((ac)^2 + (bd)^2) and area of the square is the square of that, (ac)^2 + (cd)^2, that is the sum of two squares. As remarked above, the positive numbers that are not the sum of two squares are just those that are a power of 2 times a number that leaves a remainder of 3 when divided by 4.
And every positive number is the sum of 4 squares (0 is included BTW). Three squares are a bit more complicated. A positive number fails to be a sum of 3 squares if and only if it is a power of 4 times a number that leaves a remainder of 7 when divided by 8. So the exceptions start out 7, 15, 23, 28, 31,... 
#12




Hari Seldon, you're assuming that the corners of the square must be at the lattice points. The link in the OP shows a variety of examples including one where that's not true.

#13




It is simple and my proof make it look more complicated than it is. Basically it comes down to 4m+1 can never equal 4n+3 (m, n natural)

#14




Quote:
The exact rule on when a positive integer is a sum of two squares is that this can be done if and only if each prime in its prime factorization which is of the form 3 mod 4 has an even exponent. Most of the explanation for this is contained in this old post, which deals directly with the case of primes, but also contains the unique factorization properties that allow one to properly extend to the result for general integers as stated above. Last edited by Indistinguishable; 06052017 at 07:07 PM. 


#15




Ah, I went looking for the best post to link to, but missed it. This is the best explanation for the present purposes, deriving in further detail the precise number of ways to write any positive integer as a sum of two squares (in terms of the exponents associated to primes with various remainders mod 4 in its prime factorization).
Last edited by Indistinguishable; 06052017 at 07:10 PM. 
#16




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But this triviality does not answer the OP's question, because they are not limited to squares with integral sides. 
#17




Consider lines of slope n/d in lowest terms, passing through lattice points. They are regularly spaced, with a perpendicular distance of 1/sqrt(n^2 + d^2). Thus, any two such lines have a perpendicular distance which is some integer times this.
Let n/d be the slope of one line forming one side of the square. Then this is also the slope of the line forming the opposite side of the square, and d/n is the slope of the two other sides (the spacing of such lines being at the same distance). Thus, a square formed using a line of slope n/d must have area equal to some square number divided by n^2 + d^2. If this area is a whole number, this means each prime appears in its prime factorization an even or odd number of times according as to whether said prime appears in the prime factorization of n^2 + d^2 an even or odd number of times. But by the characterization of sums of two squares as precisely those values for which each prime which is 3 mod 4 appears an even number of times in their prime factorization, this means the area must, just like n^2 + d^2, be expressible as a sum of two squares. QED. Last edited by Indistinguishable; 06052017 at 07:42 PM. 
#18




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#19




I wrote my last post without having read MikeS's post, which presents essentially the same argument slightly differently. MikeS had wondered whether it generalizes beyond 3, and indeed it does generalize completely. The key end step (in both my post and his) is knowing that if (n^2 + d^2) * area = c^2, then the area is a sum of two squares. Indeed, more generally, if A * B = C and A is a sum of two squares, then B is a sum of two squares if and only if C is a sum of two squares.
Last edited by Indistinguishable; 06052017 at 08:03 PM. Reason: A square on its own counts trivially as a sum of two squares 


#20




Expanding on that slightly…
Quote:
The direction from right to left here is trickier. Or, put another way, it's trickier to show that if B is not a sum of two squares, then neither is C. For any particular such value of B (e.g., B = 3), one will be able to elementarily reason about remainders modulo B as MikeS did to demonstrate that C is "nonhypotenuse" as well. But proving the general statement, that this approach actually works for all B, is more complicated, and where we need the less elementary theory of Gaussian prime factorization/which values are sums of two squares. Last edited by Indistinguishable; 06052017 at 08:53 PM. Reason: At least, I don't see a simpler way for that. I'd be thrilled if one could be found, though. 
#21




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Let's say a lattice line is any line which passes through two (and therefore infinitely many regularly spaced) lattice points, a latticeline square is one whose sides lie along lattice lines, and a latticepoint square is one whose corners lie on lattice points. Given any lattice line, consider the grid formed by all parallel and perpendicular lattice lines, thus carving up the plane into lots of little cells. Latticeline squares along this grid consist of, well, squares built out of these cells. But also, in particular, each cell lies in (in fact, along an edge of) some latticepoint square built out of cells (because each lattice line can be followed far enough in any direction till one ends up at one of its regularly spaced lattice points). So the area of any latticeline square is a square multiple of the corresponding cellarea, and also, the area of some latticepoint square (which is by definition a "hypotenuse number"; i.e., our slight misnomer for a sum of two squares) is a square multiple of that same cellarea. So each latticeline square's area is a hypotenuse number divided by a square and multiplied by a square. In other words, a latticeline square's area times some square is a latticepoint square's area times some square. The last fact we need is that if two whole numbers are in square ratio, and one is a hypotenuse number, then so is the other. One direction of this is trivial, while the other direction requires (as far as I can tell) the less elementary analysis (by, e.g., Gaussian prime factorization) of what kinds of integers are hypotenuse numbers. But with that, we are done. Last edited by Indistinguishable; 06052017 at 10:25 PM. 
#22




Indeed, even without the last nonelementary step, what this proves is that every latticeline square has area of the form (a^2 + b^2)/c^2, whether or not this comes out to a whole number. The last, nonelementary thing is the numbertheoretic fact that when such an expression comes out to a whole number, it must itself be a sum of two squares.
Last edited by Indistinguishable; 06052017 at 10:37 PM. 
#23




Whoops, ignore this post.
Last edited by Indistinguishable; 06062017 at 02:55 AM. 
#24




More generally, a positive rational area is achievable just in case, in its prime factorization, the exponent of 2 is at least 1, and the exponent of any prime which is 3 mod 4 is even and nonnegative.
Thus, we cannot achieve the areas 1/4, 1/8, 1/9, 1/18, etc., even though any square multiple of these which is a whole number IS achievable. The OP's example square of area 3.6, however, is achievable, since 3.6 = 2^1 * 3^2 * 5^(1), where the exponent of 2 is at least 1, the exponent of 3 (which is 3 mod 4) is nonnegative and even, and 5 (being 1 mod 4) is unconstrained in exponent. Last edited by Indistinguishable; 06062017 at 03:06 AM. Reason: Alright, alright, I'll shut up now. 


#25




I at least have enjoyed it, even if I spend more time than I should on this trying to visualize and confirm the steps to myself.

#26




Ah, I just realized a simpler way of thinking about the original problem:
Any square formed in the desired way has all its corners have rational coordinates. Which means, by choosing a common denominator for these coordinates, it is simply a square with latticepoint corners shrunk down by that denominator in each dimension. Thus, its area is (a^2 + b^2)/c^2 for some a, b, and c. [And then we invoke the number theory.] Bingobango, real simple. But it took a while to achieve this clarity… Last edited by Indistinguishable; 06062017 at 04:56 AM. Reason: This doesn't explain the 1/4 impossibility, etc, though… 
#27




Put another way, even if we didn't care about drawing the sides of the square, but just wanted to mark its corners as intersections of lines, the only restriction on square area would be that it would be of the form (a^2 + b^2)/c^2 (equivalently, of the form x^2/(y^2 + z^2) [as multiplying top and bottom by a^2 + b^2 brings one from the former form to the latter, and then symmetrically, one can go back]). And this would be by the super simple "All coordinates are rational" proof.
If we do care about drawing the squaresides, we get a little more restriction. Not just any areas of the form x^2/(y^2 + z^2) will be achievable; rather, precisely those where y and z can be chosen coprime will be achievable. And this would be by the more sophisticated "Consider cellarea by considering line distance…" proof. In both cases, the achievable whole numbers are precisely the "hypotenuse numbers", but in the former case, the achievable values in general are precisely those rationals whose lowestterms numerator and denominator are both hypotenuse numbers, while in the latter case, we add the further constraint that this denominator must further be a "primitive hypotenuse number" (one that can be written as a^2 + b^2 with a/b itself in lowestterms; equivalently, by the numbertheoretic magic we keep using from here, a hypotenuse number neither divisible by 4 nor by any nonhypotenuse factor). Last edited by Indistinguishable; 06062017 at 04:30 PM. 
#28




Then the area can be anything at all. If they are not lattice points, then I don't understand the question.

#29




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Again, look at the OP's area 3.6 UVWZ example here. Last edited by Indistinguishable; 06072017 at 08:26 PM. 


#30




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