# Diophantine fruit

I guess people may be familiar with the sorts of memes with equations involving fruit that are generally relatively trivial, while claiming something like “95% of people can’t solve this!” Sometimes this leads to amusing parodies. However someone went the whole other way and created an real, solvable instance of the problem using only natural numbers whose smallest solution is … very large.

I like the way this type of problem can actually open up a dialogue of how mathematics is really open ended, creative and unfinished. Framing a fruit equation problem as a filter that makes people either feel stupid for not understanding it (it is basically algebra, after all), or smug and superior for getting it, is extremely unhelpful. But explaining that what can seem like an easy problem actually requires a lot of work, and would stump professional mathematicians (cf Fermat’s last theorem), is a good conversation starter. It can also touch on things like how geometry is brought to bear on algebraic problems (and vice versa!), how the solution can use methods not present in the problem and so on.

So, in honour of this, I thought I take what seems to be, at time of writing, the simplest currently unsolved Diophantine equation (that is: a multivariable polynomial, and looking only for whole-number solutions), and turn it into a fruit equation. We can think of it as trying to count fruit:

Here ‘simplest’ is according to the notion of “size” defined in this MathOverflow question, basically it’s a measure of how large all the powers and coefficients of a multivariable polynomial is. There are only finitely many polynomials of a given size. The polynomial from the picture is $-x^3 + y^2 + z^2 - xyz + 5$, and has size 29. Every polynomial of size 28 and smaller has either been solved or shown to have no solutions. The idea is to see (experimentally) where the rough threshold is between equations than be solved in a more-or-less elementary way, and where really serious techniques or obstructions really kick in, of the sort outlined here.

I’m happy for people to share the above image as widely as they can. If nothing else, maybe someone will actually solve the equation above, and contribute a piece of new knowledge!

EDIT: check out the neat new preprint Fruit Diophantine Equation (arXiv:2108.02640) on this problem

## 21 thoughts on “Diophantine fruit”

1. ag24ag24 says:

Have other definitions of simplicity been discussed in the past? The arbitrary number 2 in this definition of simplicity is inelegant, since a dfferent choice leads to a different ranking of equations. I think I would prefer summing, over the terms, the absolute value of the term’s coefficient times one more than the term’s degree (so giving, for your example, 19) – but then the same inelegance arises (numbers other than 1 give different rankings). Best I have is the sum of #variables, #terms, degrees, |coefficients|, giving 27 in your case.

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1. Not sure about that. In some sense all these things are reasonably arbitrary, only that they should have roughly the same asymptotic behaviour, and only have finitely many equations of a certain complexity. There are notions of ‘height’ for more restricted formulas, see for instance https://mathoverflow.net/questions/243588/is-there-a-natural-way-to-generalize-the-so-called-bhargava-shankar-height-for-c which is defined for binary quartic forms. Ideally a notion of height should be independent of a particular parametrisation, like at the link, but for arbitrary Diophantine equations it’s not clear that there is such a thing. Maybe for homogeneous polynomials that can be seen as giving projective varieties over the integers.

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2. VW says:

6 banana, 2 apple, 3 melon.

You only have to solve a(pwr2) + 5 = m(pwr3). The rest follows from that.

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1. VW says:

I figured 3x3x3 =9. But on second thought that isn’t correct.

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2. Don’t be disheartened: literally no one in the world knows a solution to this, or even if one exists!

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3. I know this is somewhat vague, but is it suspected that there is no solution, but that’s hard to prove, or is it more open and there very well could be a solution? In other words, if you have any sense of it, what are the Bayesian priors from those who study these things as to whether a solution actually exists?

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1. I have no idea! It’s possible there is no solution, but proving this seems to be rather nontrivial.

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1. Glad I didn’t look too hard for one, then! That’s pretty cool it’s been figured out so quickly.

Thank you for this. It was very interesting.

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4. Lito Manoto says:

is this a valid solution:
Bananas=2,000,000,000
Apples=3,732,050,808
Watermelon=2,000,000,000

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1. ag24ag24 says:

No. Apart from anything else, it’s easy to show that there must be even numbers of bananas and apples and an odd number of watermelons, otherwise the two sides of the equation will have different values mod 4.

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5. ag24ag24 says:

For what it’s worth, I did a bit of exploration of the modular properties of the equation, giving the following properties that any solution must have (where x = bananas, y = apples, z = watermelons):

without loss of generality we can assume that x>0 and |y| <= x
x and y are both even
z is 1, 5 or 9 mod 10
exactly one of x, y or x-y divides by 6
if x or y divides by 6, z mod 48 is 45 if x and y both divide by 4, otherwise 9 or 33
if x-y divides by 6, z mod 144 is 125 if x and y both divide by 4, otherwise 17 or 89

Possibly those with faster hardware than mine will be able to search high enough – or maybe these constraints can be the basis of a proof?

I also had a quick look with Mathematica’s nifty FindInstance function to see which numbers other than 5 make the equation hard. It didn’t find any that are easily-proven impossible. The ones in [-1000..1000] that were undecided do not follow any pattern I can see: positive values are 5,21,33,69,93,105,129,185,237,269,273,275,281,309,321,325,349,413,465,509,521,561,573,609,635,645,715,717,721,749,761,777,785,813,829,849,857,861,905,909,989 and negatives are 67,139,159,175,183,211,219,231,355,403,455,459,475,499,519,571,579,611,643,679,707,735,763,819,859,951,999. Of these, all are 1 mod 4 except 275, 635 and 715.

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6. ag24ag24 says:

I must be missing something really obvious here, but why does k mod p = 2 imply u^2 mod p = 3? Surely it actually implies u^2 mod p = 27?

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1. ag24ag24 says:

Update: I brought this to Majumdar’s attention and it turns out that there was indeed an error, but luckily the error was only a typo in equation (4) of the paper (there should not have been a 4 before k^3), so the proof survives. What a nice proof it is too.

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1. Oh, excellent, thanks for following this up. It was late here when I posted that, and then I couldn’t spare the time to check myself.

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7. ag24ag24 says:

If of interest – a hard size-29 equation was evidently overlooked and is now under discussion here:

https://mathoverflow.net/questions/400714/can-you-solve-the-listed-smallest-open-diophantine-equations

It is 𝑥(𝑥^2+𝑦^2+1)=𝑧^3−𝑧+1. I have so far only got as far as establishing that x and z have the same sign and that x and y must both be odd (since the RHS is 6n +/-1 so does not divide by 2 or 3).

Apparently there are no hard equations of size 30.

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1. Yes, I saw that post. I don’t think the remaining size-29 equation was overlooked, rather the ones mentioned previously, including the one in the blog post, were given on math.stackexchange as examples of size-29 equations whose solubility (at the time) was unknown. I really wonder at what point the truly hard equations turn up. Maybe they have, but the techniques are known and powerful, so they don’t get a mention in this slow-burning series.

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