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February 15

Diophantine equation

Is the following Diophantine equation an instance of a well-studied class of equations, and if so, is a solution technique or impossibility proof known?

-3m^{4}+12mn^{3}-k^{2}=0

in positive integers m, n, k with n > m.

Thanks. Loraof (talk) 21:27, 15 February 2018 (UTC)[reply]

Not that I'm an expert on Diophantine equations, but I looked at this a little bit and no one else has answered yet. First note that 3|k, so letting k=3l the equation reduces to

-m^{4}+4mn^{3}=3l^{2}

or

m(4n^{3}-m^{3})=3l^{2}

Also note that a solution of the form (m, n, l)=(cm', cn', c²l') implies (m, n, l)=(m', n', l') is also a solution. So wlog we can assume the solution does not have the form (m, n, l)=(cm', cn', c²l') where c>1. This implies, after a bit of work, that m and n are relatively prime. (Basically this exploits the fact that both sides are homogeneous and the variables are separated). From there you can further restrict the form of m by considering its prime divisors. For example, if p is a prime >3, p|m but p²∤m, then p|l, p² divides the rhs, from which p|n contradicting assumption. This implies m has the form 2^a3^bu where u is a powerful number relatively prime to 6. There may be something obvious that I'm missing or an applicable theory which I'm not familiar with (algebraic number fields?) but I don't see much else to do with it. --RDBury (talk) 18:18, 17 February 2018 (UTC)[reply]

Loraof, I assume you made some brute-force, trial and error attempts at finding solutions. How high did you go? I find no solutions for 1 ≤ m < n ≤ 49796 (at which point I hit up against limits of 64-bit). There might be a probabilistic argument in favor of or against the existence of solutions, but I don't see a way to hook such a proof into this problem. -- ToE 21:36, 19 February 2018 (UTC)[reply]

Thanks to both of you. I didn’t search high for solutions — I just came across this while doodling with some math, and I figured it might impossible. I was hoping to find what the general approach would be to finding solutions to equations like this, but our article Diophantine equation is not very helpful in this regard. Loraof (talk) 15:12, 20 February 2018 (UTC)[reply]

There certainly are solutions to the given equation: a trivial family of solutions is m=n, k=±3m^2 (works for all m - of course, that is simply a scaling of the trivial solution m=n=1, k=3). Whether those are the only and how it varies with the coefficients etc. is another (tough) question though. Tigraan^{Click here to contact me} 16:44, 20 February 2018 (UTC)[reply]

One of the conditions given was n>m so it's not a solution to the problem. Not to say it's useless though since sometimes a trivial solution can lead to a non-trivial one; I don't know how that would work in this case though. --RDBury (talk) 22:02, 20 February 2018 (UTC)[reply]

Serves me well, I should try reading the question before I answer it... Tigraan^{Click here to contact me} 20:24, 21 February 2018 (UTC)[reply]

My take on this, using the hammer of p-adic valuation... Assume m,n coprime per above, and rewrite the equation as $3m^{4}*(4\left({\frac {n}{m}}\right)^{3}-1)=k^{2}$ . In that form (where the middle term is a rational number, not in general an integer), the Diophantine equation becomes a condition on the p-adic order of the middle term: there exists a k making m,n,k a solution iff $v_{p}\left({\frac {4n^{3}-m^{3}}{m^{3}}}\right)$ is odd for p=3 and even for every prime p≠3.

The trick is that for every p such that at least one of v_p(m) and v_p(n) is nonzero, this p-adic order has the same parity as m's, from which it follows that m must be three times a perfect square.

demonstration of the above claim

If v_p(m)=0 and v_p(n)>0, then the v_p of both numerator and denominator is zero (i.e. even).
For any p≠2 where v_p(m)>0, v_p(n)=0 (m and n coprime), thus v_p(4n^3)=0 and this implies $v_{p}\left(4n^{3}-m^{3}\right)=0$ . Hence $v_{p}\left({\frac {4n^{3}-m^{3}}{m^{3}}}\right)$ has the same parity as v_p(m^3), which itself is the same as v_p(m).
For the special case p=2, v_p(4n^3) is at least 2, so the v_p of the numerator is either 0 (if v_2(m)=0) or 2 (if v_2(m)>0 then v_2(m^3)>2, but then v_2(n)=0 so v_2(4n^3)=2), and the same conclusion applies.

I am not sure this helps a whole lot, but it's something (and can greatly help numerical investigation of solutions, since it significantly cuts the m's to look at). Tigraan^{Click here to contact me} 20:24, 21 February 2018 (UTC) (EDITED 16:53, 22 February 2018 (UTC))[reply]

I'm not too familiar with p-adic valuation but it looks like a useful tool here. You made a correction to the claim, but in the new version the conclusion needs to be changed a bit since the new version does not imply 3|m. Namely the conclusion now is either m is 3 times a perfect square or m is a perfect square. With the new condition on m and some more work I'm pretty sure you can show m|l and the problem becomes

4n^{3}-m^{3}=j^{2}

or

4n^{3}-m^{3}=3j^{2}

with the added condition that m is a square. This opens some new possibilities for further progress since now 4 n³ has the form j²+i² or j²+3i², both of which are well understood quadratic forms. To return to the original question, while there does seem to be progress it seems like the methods used are somewhat ad hoc, so I don't think you could claim there is a general class of equations that could be solved this way. --RDBury (talk) 21:21, 22 February 2018 (UTC)[reply]