## Diophantine Equations

Primes, divisors, arithmetic, number properties, ...
elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Diophantine Equations

I will be posting several problems related to Diophantus in separate posts so that any one may be removed without affecting the others. Have fun with them!

P.S. I've been investigating these solutions with a calculator, so any suggestions as to approximations would be welcome as well.

P.P.S. All solutions where there is any common factor between the numbers (except for the final one) are discarded. AND all solutions are integers.

One last thing: How do you pronounce his name? Is it "dee-o" or "dye-o"?
Last edited by elendiastarman on Thu Dec 25, 2008 2:23 am, edited 1 time in total.
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elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Diophantine Equations

#1:
This one actually comes from a lemma of Diophantus, but I've noticed a peculiar pattern among the solutions.

x2 + xy + y2 = z2

There seem to be a disproportional number of primes among the solutions to this equation. In fact, nearly every triple (x,y,z) has at least one prime in it. The first exception is (16,39,49), which also has the extraordinary property of having two squares among its members.
How many prime-free triples are there where 0<x<=y<=z<=l where l=1,000,000?
Are there any "square-full" triples completely composed of squares and gcd(x,y,z)=1?
Last edited by elendiastarman on Thu Dec 25, 2008 2:57 am, edited 1 time in total.
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elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Diophantine Equations

#2:
This one is an extrapolation of #1 for cubes. However, the final number is to be a square because I couldn't find any solutions where the end result was a cube.

Where 0<x<=y<=z<=n, find x,y, and z such that:

x3+y3+z3+x2y+x2z+y2x+y2z+z2x+x2y+xyz = n2

Or more simply:

(x2+y2+z2)(x+y+z)+xyz = n2
There are few 4-tuples (x,y,z,n) where n is not divisible by 10. One exception is (16,27,37,452) How many "10-less" are there where 0<x<=y<=z<=n<=1,000,000?
Last edited by elendiastarman on Thu Dec 25, 2008 2:57 am, edited 1 time in total.
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elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Diophantine Equations

#3:
This one is sort of a downgrade from #2 in a sense. This one is also a bit easier to do than #2.

Where 0<x<=y<=z, find x, y, z such that:

x2+y2+z2+xy+xz+yz = n2

Or, perhaps, more simply:

x(x+y+z)+y(y+z) + z2 = n2
Again, almost all resulting 4-tuples (x,y,z,n) have at least one prime, the first exception is (1,39,40,69). How many such prime-free 4-tuples are there where 0<x<=y<=z<=n<=1,000,000?
In almost all 4-tuples where x=1, n ends with a 9. How many 4-tuples are there where x=1, n ends with a 9, and x<=y<=z<=n<=1,000,000?
Last edited by elendiastarman on Thu Jan 01, 2009 3:34 pm, edited 1 time in total.
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elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Diophantine Equations

What? No responses? Weird...

Before I got distracted about a week ago, I was working on the idea of using the fact that all squares leave a remainder of 0 or 1 when divided by 4. That eliminated about half of all possible cases (of values [mod 4]) for x, y, and z. Of course, this was before I saw the simplified form, so... . Time to work some more...
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uws8505
Posts: 58
Joined: Tue Sep 30, 2008 3:13 pm
Location: South Korea

### Re: Diophantine Equations

elendiastarman wrote:#3:
...
Again, almost all resulting 4-tuples (x,y,z,n) have at least one prime, the first exception is (1,39,40,59). How many such prime-free 4-tuples are there where 0<x<=y<=z<=n<=1,000,000?
I think you missed that 59 is a prime. Or are you considering only primality of x, y, z but not n?
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daniel.is.fischer
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### Re: Diophantine Equations

Re #1: There are 137,842 primitive triples with z [le] 1,000,000 and 98,256 of them are prime-free. I strongly suspect that no such triple can consist of three squares, but so far I have no proof.
Il faut respecter la montagne -- c'est pourquoi les gypa&egrave;tes sont l&agrave;.

elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Diophantine Equations

Whoopsie, that 59 is supposed to be 69, which is composite. Sorry!

@daniel.is.fischer:
Well, since we already know that x4+y4 &ne; z4, all that needs to be proved is that adding x2y2 doesn't change anything....which, I suppose, is the hard part...
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Eigenray
Posts: 62
Joined: Mon Jul 14, 2008 5:20 pm

### Re: Diophantine Equations

It should be possible to prove by descent that there are no primitive solutions to
N(x-wy) = x2 + xy + y2 = z2
with x and y both squares. Here w is a primitive cube root of unity and N denotes the field norm. If a prime p=2 mod 3 divided z, then it would have to divide x-wy and therefore both x and y. For a rational prime p=1 mod 3, we can't have both primes lying over p dividing x-wy, or else again p would divide x,y. And looking mod 9 shows x-wy can't be divisible by the prime over 3.

Since N(x-wy) is a square, it follows that x-wy is itself a unit times a square:

x - wy = (-w)k (a - wb)2, with x and y squares, k=0 or 1.

Case 0: x = a2 - b2, y = 2ab + b2.
By the first equation we have a = m2+n2, and b either 2mn or m2-n2.
-Case (i): If b=2mn, then y = 4mn(m2+mn+n2) is a square.
Since (m,n) are relatively prime, m, n, and m2+mn+n2 must all be squares. We've arrived at a smaller solution to the original equation.
-Case (ii): If b=m2-n2, then y = (m-n)(m+n)(3m2+n2). Making the change of variables u = m-n, v = m+n gives y = uv(u2+uv+v2), which also gives a smaller solution.

Case 1: x = -b(2a+b), y = a(a+2b).
WLOG a>0, b<0. Looking mod 4, a must be odd and b even. Since (a,b)=1, a is a square and -b is either a square or twice a square.
-Case (i): a = A2, b = -B2, A2-2B2 = u2, 2A2-B2 = v2. Then 2v2-u2 = 3A2 = 3 mod 8, which has no solutions.
-Case (ii): a = A2, b = -2B2, A2-4B2 = u2, A2-B2=v2. Setting A = m2+n2, 2B = 2mn, the last equation is equivalent to m2+mn+n2 and m2-mn+n2 both being squares. Hmm. Actually I'm stuck here.

Well, it's "reduced" to showing the only points on
y2 = a4 + 6a3b + 13a2b2 + 6ab3 + b4
or
y2 = a4 + 2a3b + a2b2 - 6ab3 + 3b4
with (a,b)=1 are the trivial ones with -2 [le] a,b [le] 2.

Eigenray
Posts: 62
Joined: Mon Jul 14, 2008 5:20 pm

### Re: Diophantine Equations

Consider rational solutions to
z2 = x4 + x2y2 + y4.
If y=0 then z = +/- x2. Otherwise, we may replace z := z/y2, x := x/y, to assume y=1:
z2 = x4 + x2 + 1.
This is an elliptic curve. Setting
X = 2x2+2z+1, Y = 2x(2x2+2z+1)
gives
Y2 = X3 - 2X2 - 3X = X(X-3)(X+1),
which can be put into a program like Magma:

Code: Select all

E:=EllipticCurve([0,-2,0,-3,0]);
RankBounds(E);

The rank is 0, and the torsion points are easily seen to be (0,0), (3,0), (-1,0). These give only (x,z) = (0, -1), (0, 1) on the original curve. So there are no solutions with x and y non-zero.

As for how Magma is able to compute the rank, it is most likely via complete 2-descent. See Proposition 1.4 in Silverman's Arithmetic of Elliptic Curves.

For our curve E, let K = { +/- 1, +/- 2, +/- 3, +/- 6} (the integers mod squares and odd primes of good reduction). The rank of E/(2E), which is 2 more than the rank of E, is the dimension of the image of E/(2E) under a certain homomorphism into K x K, and a point (u,v) (other than (1, +/-1), (1, +/-3)) is in the image iff the system
ux2 - vy2 = 1
ux2 - uvz2 = 4
has a solution. We can eliminate any pairs with u < 0, leaving 32 possibilities. But since the image is a subgroup of K x K, we can actually eliminate pairs 4 at a time. So there are really only 7 curves to check.

Edit: Actually they are all easy to check. Put (x,y,z) = (a/n, b/n, c/n), and consider
ua2 - vb2 = n2
ua2 - uvc2 = 4n2.

If 3 divides u but not v, then we get 3 divides n, b, a, and then c, a contradiction. This kills (3,1), (3,2), (6,1) and (6,2), and by the fact that the image is a subgroup, everything with u=3 or 6. For (2,1), 2 divides n,b,a,c. For (2,2), 2 divides a,n,b,c. And for (1,2), 2 divides a,n,b,c. This kills everything except the images of the four torsion points, so the curve has rank 0, as claimed.

karlo
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### Re: Diophantine Equations

One can show that all solutions in natural numbers to x^4+x^2y^2+y^4=z^2 are (k,0,k^2) and (0,k,k^2).

Solution:
We can assume (x,y)=1. Then x and y have different parity. Wlog y is odd and minimal with this property. Rewrite the equation as:
4z^2-(2^x^2+y^2)^2=3y^4 or (2z+2x^2+y^2)(2z+2x^2+y^2)=3y^4. Let d=(2z+2x^2+y^2,2z-2x^2-y^2). Then d is odd and d divides z and 2x^2+y^2. As a result d|3y^4. If d|y^2, we get (x,y)>1, so d|3y. Let p a prime factor of d. If p>3 then p|y, so (x,y)>=p, a contradiction. If p=3, then we get (x,y)>=3. As a result d has to be 1.
Therefore, 2z+2x^2+y^2=a^4, 2z-2x^2-y^2=3b^4, y=ab or 2z+2x^2+y^2=3a^4, 2z-2x^2-y^2=b^4, y=ab, where a,b are both odd and (a,b)=1.
Case 1: 2z+2x^2+y^2=a^4, 2z-2x^2-y^2=3b^4, y=ab => 4x^2 is -4 mod 16, a contradiction.
Case 2: 2z+2x^2+y^2=3a^4, 2z-2x^2-y^2=b^4, y=ab => 4x^2=(a^2-b^2)(3a^2+b^2) => a^2-b^2=c^2 and 3a^2+b^2=4e^2. Therefore, a=p^2+q^2, b=p^2-q^2, and we get p^4+p^2q^2+q^4=e^2, which contradicts the fact that y is minimal.
As a result, y=a=b=1 and x=0 Which gives the solution (0,1,1). By symetry, we also get the solution (1,0,1). The conclusion follows.

Note: solution not due to me.

boris-nn
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### Re: Диофантовы Уравнения

We prove that in the particular case of the system of equations has no solution in integers.(arXiv:1305.3380)
m2+n2=p2
m2+k2=q2
n2+k2=z2
n2+q2=g2
Also in this article is a description of the method by which you can write a simple algorithm for finding all the bricks Euler (cuboid in which one element (edge or any diagonal) non-integer.