A tricky math problem: restricting a in the equation a(b^2-1)=c^2+d^2 to force |b|=1?

Hey, this is what we’re studying in class right now! But I don’t know, sorry.

I got a headache just from reading your question’s details. I am a math idiot for the most part and I couldn’t resist making this comment. Sorry if it appears out of place here.

removed by me

Here comes the gene pool to the rescue. Such a family, nu?

Won’t having |b|=1 mean the whole left side of the equation will be zero, thereby forcing c=0 and d=0, and making a unbound. That doesn’t seem like a tricky math puzzle, so I think I’m missing something.

I want b=1. But i can only put conditions on a. I’m tryting to prevent other solutions.

@finkelitis Ah, I see. So, for example, if c and d were cubed instead of squared, you could use Fermat to say if a were a cube, b must equal 1. That is hard.

edit: Is that the same as asking, “What kinds of numbers can’t be the sum of two squares?”

Homework. Weee…

@loser fwiw, @finkelitis (my brother) is in a math ph. d program, and, as he states, this is a tricky math problem, not homework.

To everyone: this is a hard, and possibly unsolved problem. Don’t feel like you’re stupid if you don’t get it. I’m kind of fishing for the professional amateur mathematician set on fluther. You have to be a little specialized for this one.

@ben there is a good answer to your question “What kinds of numbers can and can’t be written as the sum of two square?” The answer is: decompose the number as a product of primes. If all the odd primes are 1 mod 4, then the number can be written as a sum of squares. If any are 3 mod 4, then it can’t be written as a sum of squares.

Unfortunately, that doesn’t help me here because if b is odd, the left side is 0 mod 4, and I get no further information.

Also, even in the case that you did have the situation you described, you still couldn’t use Fermat, because you still have b^-1 floating around to screw things up.

In fact, what I’m really looking for is a number theorist to tell me if this problem has already been solved, and if it hasn’t been, how hard it is.

finkelitis- This might not help but I think you can pretty easily give a solution provided the following statement is true:

For any two relatively prime positive integers a and b there is two integers u and v such that a*(u^2+v^2)+b is a prime number. (Or at least it is the sum of two squares.)

This is a generalization of Dirichlet’s Theorem and I have no idea if it is true or not. Just thought I might share my random thought!

If this statement is true I think I can show that any number a can be written as (c^2+d^2)/(b^2–1).

The reason this statement proves what you want is as follows:

Lets say p1,...,pk are those primes in a that their exponent in the prime factorization of a is odd. Take b = 8 p1…pk (u^2+v^2)+1

Then if you evaluate b^2–1 you get: 16 p1…pk (u^2+v^2) (4 p1…pk (u^2+v^2)+1).

To make a*(b^2–1) the sum of two integers “all” we need is to be able to write (4 p1…pk (u^2+v^2)+1) as the sum of two integers. Again this might not help. Just my (probably useless) 2 cents! :)

>> To make a*(b^2–1) the sum of two integers “all” we need is to be able to write (4 p1…pk (u^2+v^2)+1) as the sum of two integers.

Make that: as the sum of two squares.

I just walked in and all I can say is “who farted?” Oh…I see…just math. Well, math, saurkraut and farts all smell the same to me.

sorry for popping in…I’ll take Marg A. Rita to the next room and make sure she is comfortable and then I’ll stop back to help solve this

@ben I barely passed algebra! The memories!!! No offense, I swear!!!

As I said before for a fixed a we are looking for u and v such that a(u^2+v^2)+1 can be written as a sum of two squares. So we get a(u^2+v^2) = x^2+y^2–1

If you put x=0 we are basically going to get an equality which is equivalent to the equality in the question.

This new equation gives us more freedom, because we have more variables. I would like to somehow add more variables and finally somehow be able to use the Four Square Theorem, but I don’t know how.

P.S. I am guessing there is no such a. and this is basically what I am trying to prove.

finkelitis -

>> If all the odd primes are 1 mod 4, then the number can be written as a sum of squares. If any are 3 mod 4, then it can’t be written as a sum of squares.

I think you may have made a mistake on your second sentence. The primes that are 3 mod 4 should have an even exponent. They may exist but their exponent should be even, if they do.

@BonusQuestion—You’re right in your last comment… I need to specify that the number should be square free, which is a reasonable assumption.

@BonusQuestion—Just went through your reasoning. It’s an interesting tack. I don’t know the answer to the reformulated question either though. Thanks for the response though—hitting you with some lurve.

If you’re going to derail, whisper your comments, people. Like this:

Honestly @finkelitis, do we have to do your homework for you? Isn’t it OBvious? Just take occum’s razor and apply De morgan’s law, after which you the ex post facto reduction in terms and then you simply account for n + 1. QED

@andrew: Back to remedial math for you, my pretty. (Remember, too, that I have more than one nephew and guard all their backs.)

I asked the best mathematician I know:

This is a difficult math question. The solution, if it exists, probably involves residues and modular arithmetic, and maybe some clever substitutions… right now it’s best to look at individual cases. For example,

1) a(b^2 – 1 ) = c^2 + d^2

we want to see if there are conditions for a that make a solution to this equation impossible except for the trivial solution (b = +- 1 => c^2 + d^2 = 0 => c=d=0)

so

solve 1) for b^2 =>

2) b^2 = (c^2 + d^2)/a + 1 = c^2 / a + d^2 / a + 1

the author of the question mentioned the special case a is a perfect square a = p^2 where d = 0

then b^2 = c^2 / p^2 + 1 = (c / p)^2 + 1

so c/p < b and they must both be integers…

therefore c/p = b-k for some positive integer k

BUT THEN:

b^2 = (b-k)^2 + 1 = b^2 -2bk+k^2 +1 =>

0 = k^2 – 2bk + 1 => 2bk = k^2 + 1 =>

b = (k^2 + 1) / 2k = k / 2 + 1 / 2k

if k > 1 then k / 2 + 1 / 2k cannot be an integer.

if k = 1 then… b = 1 / 2 + 1 / 2 = 1

b = 1… the condition we wanted.

So thus if a is a perfect square and d = 0 then b = 1.

Also, for a simple nontrivial example, if a = 1 then

1 * ( 3^2 – 1) = 2^2 + 2^2

is a possible solution.

I would have to guess it is probably harder than a problem that can be answered by elementary methods.

If I wanted to attack this problem I would have looked at the proof of Dirichlet’s Theorem. Using some analytic number theory they evaluate the analytic density of primes of the form an+d and their density is 1/phi(a).

If you go into the proof of Dirichlet it may be possible that you are able to restrict n to numbers that are sums of two squares.

FWIW, any a that is the sum of two squares can be written as (b^2+c^2)/(d^2–1). The reason is that you simply replace d=3 and you get (b^2+c^2)/8. But we already know a is written as the sum of two squares. So 8*a can be written as the sum of two squares which means a is the sum of two squares divided by 8.

Hence a can NOT be the sum of two squares. (e.g. a cannot be a square.) So, forget about simple conditions on a (like a is a perfect square or cube, etc.) You should be looking into its prime factors of the form 3 mod 4. That should probably give us the condition we are looking for. Although I am thinking the condition does not exist

I suspect that this question may be equivalent to the statement (or a similar one) that I posted above.

I am thinking the next step would probably be to see if for any prime p of the form 3 mod 4 we can find a, b, and c such that p = (a^2+b^2)/(c^2–1).

What I would personally do: I would send an email to a number theorist and ask.

For primes, I can now prove the “equivalency statement” that I claimed above.

More precisely, I am claiming that the following two statements are equivalent.

1) For any prime p of the form 3 mod 4, p can be written as (a^2+b^2)/(c^2–1).

2) For any prime p of the form 3 mod 4, there is a number n which is a sum of two squares such that at least one of these numbers is also a sum of two squares: pn+1, pn+2, pn-1, pn-2

1 => 2: p(c^2–1) is the sum of two squares. Since p is 3 mod 4, c^2–1 is divisible by p. So c is in form of pn+1 or pn-1. (Since p is prime.)

case i) c=pn+1

p(c^2–1) = p pn (pn+2) = p^2 n(pn+2)

This is the sum of two squares. Therefore n(pn+2) is also the sum of two squares. But you can easily see that g.c.d(n, pn+2) = 1 or 2.

If their gcd is 1 then both should be sums of squares which means n and pn+2 are sums of two squares. which proves (2).

If their gcd is 2 then n/2 and pn/2+1 are coprime and hence they are both sums of squares. Which proves (2).

case ii) c=pn-1

p(c^2–1) = ppn(pn-2)

This is the sum of two squares. Again similar to above argument n(pn-2) is a sum of two squares. gcd(n, pn-2) = 1 or 2 and with the similar argument you can say either n and pn-2 are both sums of two squares OR n/2 and pn/2–1 are sums of two squares.

2 => 1: Lest say n is a sum of two saquares and pn-1 is also the sum of two squares. Set c= 2pn-1. We have p(c^2–1) = p 2pn(2pn-2) = 4p^2 n (pn-1) but n and pn-1 are sums of two squares so this is also a sum of two squares.
The other 3 cases are similar.

I think I can give a complete proof now. Namely, I show for any a there is 3 numbers b, c, and d such that |b| is not one and a (b^2–1) = c^2+d^2.

If a is a perfect square, say a=x^2, then take b=3, c=2x, d=2x.

a(b^2–1) = 8a = 8x^2 = 4x^2+4x^2 = c^2 + d^2

Which means we have found a nontrivial solution.

Now assume a is not a perfect square.

By a result from Lagrange for Pell’s equation there is two positive natural numbers x and y such that x^2 – a y^2 =1.

Now set: b = x, c = ay, d=0

a (b^2–1) = a ( x^2 -1) = a ay^2 = c^2 = c^2 + d^2.

And we have found a nontrivial solution.

um… yeah, bonus question beat me to the answer. What he said.

+5 for me and ben for creating fluther. and BonusQuestion (OBviously)

You’re totally right, BonusQuestion. And you’ve incidentally answered another question that was on my mind (about the d=0 case).

Thanks for the insight!

You’re welcome. Glad it helped.

For those who are interested in the solution of Pell’s equation, there is a rather elementary method for finding those solutions. For details check this pdf out.