## Maths challenge 1

Posted by Chris on April 28, 2011 – 2:27 pm

Solve x² + y² + z² = 2xyz, where x,y,z are integers > 0 (i.e. natural numbers).

Addendum: In fact that isn’t possible to do. So I’ve ended up proving that

x² + y² + z² = 2n xyz cannot be solved for n,x,y,z in natural numbers.

April 28th, 2011 at 4:57 pm

I have number crunched for x,y,z up to 500 without joy. But unless the source site has goofed or it’s a trick question, it might be soluble.

April 29th, 2011 at 1:13 am

Hi Chris. This equation doesn’t have any integer solutions! I will not spoil the fun for the others and not explain why (it is not difficult at all, though – I haven’t written a single line while “solving” it). I am still interested in my question from post #7 of “Integer Divisors” which is much more general

April 29th, 2011 at 4:59 am

Hi slavy, my guts tell me you’re right (after some futile attempts at solving it). I can only come up with trick answers like x = y = √2, then 2 + 2 + z² = 4z => z = 2, i.e. the LHS and te RHS are integers, not all the unknowns.

The source site must have goofed, it is very unlikely that it was a trick question.

I’d forgotten that a related problem had been posted. Here’s a link to your question: http://trickofmind.com/?p=721#comment-4186 => for what values of r does x² + y² + z² = rxyz have an integer solution? We know that r = 1 and r = 3 work.

April 29th, 2011 at 7:48 am

Hi Chris. I am certain that the answer is no – I can prove it easily. Furthermore, it follows that for all even r’s the answer remains negative. Due to arguments stated in the previous problem, a necessary condition for r=r1*r2 to admit integer solution is that both r1 and r2 admit, as well! This explains why I know that even r’s don’t work and that the problem is interesting mainly for prime factors r (it is more convinient to denote it by “p” in this case)…

April 30th, 2011 at 7:37 am

Hi slavy. I agree with your findings, and I can prove it too, and I can do it in my head .

So there’s the challenge folks: prove that x² + y² + z² = 2nxyz cannot be solved with n,x,y,z as natural numbers (i.e. 1,2,3,…).

I used the properties of square numbers, mod 4, that slavy showed in a camel sharing problem a few months back.

May 1st, 2011 at 8:44 am

Any integer m can be written as m = 2n + r, where n is any integer and r = 0 or 1. Then m² = 4n² + 4nr + r² and so m² = r² (mod 2 and 4). Therefore m² = 0 (mod 2 and 4) if m is even and m² = 1 (mod 2 and 4) if m is odd.

The equation can be written x² + y² + z² = 0 (mod 2) , but that means that one of x,y,z is even and two are odd, or all three are even. Either way, we then have that 2xyz = 0 (mod 4). So we must have that x² + y² + z² = 0 (mod 4). That can only be satisfied with x,y,z all even. To save introducing new variables, replace x,y,z with 2x,2y,2z and substitute into the original equation => x² + y² + z² = 4nxyz. But 4nxyz = 0 (mod 4), and again we have that (the new) x,y,z are all even. Once again, replace x,y,z with 2x,2y,2z to get x² + y² + z² = 8nxyz. It seems that we could keep this up forever. However, x,y,z are finite, so we must eventually run out of factors of 2 for at least one of them, x say, and then that final x will be odd – but that cannot be a solution. That’s all folks.

May 1st, 2011 at 10:31 am

Very good, Chris I have nothing to add here. For the general case – if x^2+y^2+z^2=pxyz is considered as a quadratic equation with respect to one of the variables (let’s say z), then the question of integral solution is equivalent to checking if the discriminant of the equation is a perfect square, i.e., to solving (pxy)^2=(2x)^2+(2y)^2+D^2. For the latter, we can safely assume that the greatest common divisor of (x,y)=1 and thus to work with the explicit formula for Pythagorean boxes (or quadruples). I don’t have time to look into it deeper, so I still don’t know if this leads to something or not… Some volunteers may write a computer routine to check the cases p=5 and p=7 at least for small (x,y,z). It is not possible to use simple mod p arguments, as in the case p=2, because we have many many cases, e.g., (x=5x, y=25y+1, z=25z+7), or (x=5x, y=25y+3, z=25z+4), etc.

May 1st, 2011 at 12:59 pm

Hi slavy. Thanks. I don’t know if I would have succeeded if I didn’t know that mod 4 trick. The mod 2 bit saved a lot of arm waving. I’m guessing that’s pretty much how you got there. Although it took a fair amount of writing, I had done it all in my head a few days ago.

I used a computer to check up to p = 17 (i.e. p = 5,7,9,11,13,15,17) and x,y,z up to 200; no solutions were found. So I conjecture that p = 1 and p = 3 are the only allowable values. I decided not to risk assuming that p was prime.

I’ve been away for the weekend. I’ll test with more values this evening.

May 1st, 2011 at 5:37 pm

I’ve number crunched (nxyz) for n = 1 to 51 (odd values only) and x,y,z = 1 to 350 – no solutions found.

Meanwhile, I’ve started to think about the more general case x² + y² = z² = nxyz.

If x = y = z ≠ 0,then, 3x² = nx³ => 3 = nx. Then either n = 1 and x = 3 or n = 3 and x = 1.

if y = z ≠ x (and none are 0) then x² + 2y² = nxy² => x² – ny²x + 2y² = 0

=> 2x = ny² ± √((ny²)² – 8y²) =>2x = ny² ± y√((ny)² – 8). So we need

(ny)² – 8 to be a perfect square = m² say => (ny – m)(ny + m) = 8 = 1*8 = 2*4. That can only be satisfied with ny = 3 and m = 1. So we must have n=1 and y = 3 or n = 3 and y = 1. That leads to n = 3, y = z = 1, x = 2 and reproduces the previous result.

If x,y,z are all different, we get 2x = nyz ± √((nyz)² -4(y²+z²)) and so we need (nyz)² -4(y²+z²) = k². That discriminant doesn’t obviously lend itself nicely to comparison with Pythagorean triples.

However, n = 1 and 3 give solutions with x=y=z, y = z ≠ x and x,y,z all different, but that involved using a computer.

I am going to have another bash using modulo arithmetic, but in view of slavy’s comments, I shan’t spend too much time on it. Excercises in futility are not my cup of tea.

May 2nd, 2011 at 2:58 am

I’ve now number crunched nxyz for n = 1 to 161 (odd values only), x,y = 1 to 500, z to 1000. I only had success for n = 1 and n =3 (as expected). I’ve now stopped the program.

n = 1 => {3,3,3}, {3,3,6}, {3,6,15}, {3,15,39}, {3,39,102}

{3,102,267}, {3,267,699}, {6,15,87, [6,87,507}, {15,39,582}

n = 3 => {1,1,1}, {1,1,2}, {1,2,5}, {1,5,13}, {1,13,34},{1,34,89}, {1,89,233},

{1,233,610}, {2,5,29}, {2,29,169}, {2,169,985}, {5,13,194}, {5,29,433}

Obviously there are (infinitely) many more.

I found a nice page on Pythagorean triples:

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html

Here’s one: Take any two natural numbers that differ by 2, e.g. 11 and 13. then 1/11 + 1/13 = (11+13)/(11*13) = 24/143, then 24 and 143 are the lengths of two of the sides of a right angled triangle. In this case the third side is √(24² + 143²) = 145.

Here’s another: Take any two fractions (including integers) whose product is 2. e.g. 1/5 and 10. Add 2 to each => 11/5 and 12, multiply both by the lcm of the denominators (5 in this case) to get 11 and 60. These are the short sides of a right-angled triangle. The hypoteneuse in this case is √(11² + 60²) = 61.

I’ve skipped a lot of details in those examples.

May 2nd, 2011 at 3:18 pm

Hi, Chris. Thanks for the simulations I gave up on the problem and asked some friends for help. When p=3 the equation is known in the literature as the Markov’s equation http://en.wikipedia.org/wiki/Markov_number

(A. Markov since there are several famous Russian mathematicians with the same family name). I didn’t spend much time on the details, but in the following short paper http://www.springerlink.com/content/r158635321r6305w/ there is a theorem (the Corollary) which confirms that for p>3 there is no hope for an integral solution. The author considers a much more general equation though and the results are highly nontrivial, so I don’t know if there is an easy way to prove it…

May 2nd, 2011 at 4:53 pm

Hi slavy. I think you mathematicians are all a little bonkers

But having said that, I had fleetingly considered a²+b²+c²+d² = nabcd, but that moment passed uneventfully.

May 3rd, 2011 at 1:42 am

Hi Chris. I agree By the way, the general formula in the above paper covers all the extensions you can have in mind. The result is that whenever the coefficient n from the right-hand-side is bigger than the sum of all the coefficients from the left-hand-side we never have integral solutions. So here, for n>4 we can’t hope for anything, n=4 and n=1 admits infinitely many solutions (start with the solution (1,1,1,1), resp., (2,2,2,2) and march upwards), while for n=2 and n=3 we need additional computations (or maybe not – I didn’t think of it)

May 3rd, 2011 at 5:09 am

Hi slavy. Unfortunately, I can only access the first page of that second link. They want money to see more/download the pdf. But even the first link had led to quite involved looking stuff.

I’ve slightly tightened up my argument in post 6. There’s less hopping about at the beginning.

Anyway, at least I found some nice (albeit useless ) stuff on Pythagorean triples and of course once more was able to appreciate the extraordinary utility of modulo arithmetic.

May 3rd, 2011 at 8:48 am

Hi Chris. I have access to the whole paper from the university, so if you are interested I can send it to you. it is only two pages overall with 3 theorems and one corollary, which is basically the answer of the question and absolutely no proofs

May 3rd, 2011 at 11:25 am

Hi slavy. Yes, please send it.

May 3rd, 2011 at 3:12 pm

ОК – I will send it tomorrow when I go to the university to the e-mail address I have from a previous correspondence.

May 6th, 2011 at 8:17 am

Behind the scenes, slavy has mentioned looking at the problem in mod 3. I haven’t completely finished doing that, but I have enough to make it worth publishing.

Any integer m can be written as 3n + r, where r = -1,0,1. So m² = r² (mod 3) => m² = 0,1 (mod 3). 0=> divisible by 3 and 1 => not divisible by 3.

Now consider x² + y² + z² = nxyz. There are two possibilities, none of x,y,z is divisible by 3, or at least one is. If none are divisible by 3, then x² + y² + z² = 1+1+1 = 0 mod 3. And nxyz = n(±1)(±1)(±1) = ±n (mod 3). But that must = 0 (mod 3), so n must be divisible by 3 if none of x,y,z are (if the equation is soluble).

If at least one of x,y,z is divisible by 3, then nxyz = 0 (mod 3) and then we must have that all three of x,y,z is divisible by 3, but then n has no restriction on it.

Considering the case of x,y,z all divisible by 3. We can incorporate that by replacing x,y,z with 3x,3y,3z to get x² + y² + z² = 3nxyz. In fact, it is pretty obvious that if the original x (say) had exactly k factors of 3, then y and z must as well. So we can rewrite the original equation as x² + y² + z² = 3

^{k}nxyz where none of (the final) x,y,z have a factor of 3. It also seems sensible to observe that n can be replaced with 3^{j}n to arrive at the (final) equation x² + y² + z² = 3^{j+k}nxyz where none of n,x,y,z is divisible by 3. NB if the original n wasn’t divisible by 3, then j = 0, and if x (and hence y,z) wasn’t divisible by 3, then k = 0.From previous considerations, we “know” that j+k = 0 or 1 for a solution set. I might add more later.

May 6th, 2011 at 9:20 am

Just a remark – Chris you take not (3x,3y,3z) but (x/3,y/3,z/3) which is an integral solution, since all x,y,z are divisible by 3. Then we have (x/3)^2+(y/3)^2+(z/3)^2=3n(x/3)(y/3)(z/3) and analogously for the higher powers if we have that x=3^kx, y=3^ky, z=3^kz…

May 6th, 2011 at 9:36 am

Hi slavy. I did it as, if x is divisible by 3 then x = 3u (say) (and similarly for y and z). Then we get (3u)² + (3v)² + (3w)² = n (3u)(3v)(3w) => u² + v² + w² = 3nuvw. But then I replace u,v,w with a redefined x,y,z to save extra variables. i.e. I’m removing the factors of 3 from x,y,z to get a (relatively) primitive solution.

May 21st, 2011 at 6:09 pm

x² + y² + z² = nxyz. Without loss of generality, if gcd(y,z) = f and is the highest gcd amongst all the pairs, then replacing y and z with fy and fz => x² + f²y² + f²z² = x² = nf²xyz = 0 (mod f²). So x must also have a factor f (and no other factor common with y and z, otherwise f wouldn’t have been the highest gcd amongst the pairs). Replacing x with fx gives: x² + y² + z² = nfxyz where x,y,z are fully (pairwise) co-prime (or a pair or all three are equal). If e.g. y = z = f, then we get x² + 2 = nfx, or if x = y = z = f => 3 = nf. I believe I’ve covered those in post 9.

I’m chipping away, but still can’t see the final move.

June 2nd, 2011 at 6:14 pm

In stumbling around for new problem, I found the following (reflection of my last post), but applied to Fermat’s last theorem.

If x

^{n}+ y^{n}= z^{n}has a solution then there is a solution for which x,y,z are all co-prime to each other. Rewrite the equation as x^{n}+ y^{n}– z^{n}= 0. It is obvious that if gcd(y,z) = d, say, then letting Y = y/d and Z = z/d, we get x^{n}+ (dY)^{n}– (dZ)^{n}= 0 = x^{n}(mod d^{n}). Therefore x must also have a factor d and so we have X = x/d =>X

^{n}+ Y^{n}= Z^{n}.This applies to any 2 pairs, so altogether we can eliminate all common factors.

June 2nd, 2011 at 8:17 pm

… also, for the co-prime case, if Fermat’s last theorem was false, then exactly one of x,y,z would have to be even – no two (or three) can be even, as shown in my last post.

June 13th, 2013 at 7:47 am

Hi every body,

Could you please help me to resolve this problem, i.e find solutions;

Consider a number m in N\{0}, with m is odd,

Résolve the equation x²+y²+z² = xz mod[m], ?(x,y,z)

June 14th, 2013 at 3:16 am

Hi rabi. I suspect that’s a difficult problem.

Even a simple problem like e.g. x

^{2}= 3 (mod m), m prime, requires the use of quadratic reciprocity to find allowable values of m. So I don’t feel very hopeful about finding a solution.Allowing m to be composite, rather than a prime mod, is less easy as have to use the Jacobi symbol rather than the Legendre symbol.

Of course, I might be thinking about it in completely the wrong way.

June 14th, 2013 at 4:43 am

Hi rabi. The equation definitely has modular solutions – I checked by number-crunching. Unless I think of something amazingly clever, I’m pretty sure that finding a general solution is beyond by abilities.

Do you have a clue about how to make any progress? Why do you want to solve the equation?

June 14th, 2013 at 6:10 am

Hi Chris,

Yes I agre that is not a easy problem, even if we take m as a prime. So now, if we consider the equation :

x² + y² + z² = 2xz in N, is it possible to find a solutions in N^3\{0,0,0}?

We know that the equation (*) x²+y²+z²=xyz have solutions, and if (x,y,z) is one of them we can calculate another solution by doing (y,z,yz-x)…

Now, about my equation, if I take y=2 then I have the forme, so the quation now is; Is there a solution of (*) witch can be like this (2,b,c)?

June 14th, 2013 at 6:21 am

Oh!! excuse me chris,

I seed that in challenge 1, this equation have no solutions in N.

June 14th, 2013 at 11:46 am

Although x² + y² + z² = 2xyz can’t be solved in the non-zero integers, it doesn’t follow that it can’t be solved as a modulo problem.

e.g. (x,y,z) = (2,2,2) (mod 4)

(1,1,3), (1,1,4), (1,2,4), (3,4,4), (4,4,4) (mod 5)

(I think that’s all the unique solutions mod 5).

June 14th, 2013 at 3:06 pm

I’ve just re-read slavy’s challenge in post 7.

That’s to find the solutions of x

^{2}+y^{2}+z^{2}= pxyzThe equation certainly has solutions mod 5. e.g. {p,x,y,z} = {1,1,2,2},{1,1,3,3},

{1,2,2,3},{1,2,3,4},{1,3,3,3},{2,1,1,3},{2,1,1,4},{2,1,2,4},{2,3,4,4},{3,1,1,1},

{3,1,1,2},{3,1,3,4},{3,1,4,4},{3,2,2,4}

The list goes on, but I shan’t. There are also many cases with 0’s in them.

There’s also a lot of mod 7 solutions. Most of them don’t have a 0 in them.

I know, via slavy, that we need p = 1 or 3, to have diophantine solutions.

June 25th, 2013 at 5:52 am

Thank you Chris.

June 25th, 2013 at 6:29 am

Hi Rabi. My pleasure. I’m always pleased to revisit old problems. Sometimes I learn/discover something new when I do that.