In what field does $\pi^2=2$, where $π$ is the ratio of the circumference of a circle to its diameter?

Remember to think outside the box and that I am looking for a complete answer.


The above is to be taken more literally than figuratively and vice versa respectively.

  • 6
    $\begingroup$ When you set $\pi = \sqrt{2}$, I'm guessing. $\endgroup$ – Joe Z. Apr 13 '15 at 6:39
  • 6
    $\begingroup$ If by π you mean 3.14159... then the answer is: never. Outside of the box, there are plenty of answers. It looks like a "guess what I am thinking" riddle. $\endgroup$ – Florian F Apr 13 '15 at 7:37
  • 2
    $\begingroup$ @FlorianF, by $\pi$ I mean the ratio of the circumference of a circle to its diameter. I have restructured it slightly to hopefully avoid looking like a "guess what I am thinking" riddle. $\endgroup$ – Joel Bosveld Apr 13 '15 at 8:23
  • 3
    $\begingroup$ I think maybe we should look at the π symbol as a word 'pie'. Piesquared is a pizza place in Canada I think, but I don't know why it equals two. $\endgroup$ – Zikato Apr 13 '15 at 8:55
  • 2
    $\begingroup$ Any chance of a hint or solution for this? Have any of the answerers below got the correct answer? $\endgroup$ – Rand al'Thor Apr 26 '15 at 15:33

10 Answers 10


Answer 1

The field is cooking.
$\pi^2$ milliliters = 2 teaspoons

Answer 2

The field spherical geometry.
For a small circle of radius subtending angle $\theta = 2.010311...$ radians at the centre of the sphere, the ratio between the circumference and the diameter measured on the surface of the sphere is $\sqrt2$. In this precise case we could say $\pi^2=2$.

  • 3
    $\begingroup$ Given that a teaspoon is used only in cooking, an error of 0.12% is more than acceptable. I.e 2 teaspoons +/- 1% are still 2 teaspoons. $\endgroup$ – Florian F Apr 13 '15 at 10:49
  • $\begingroup$ @FlorianF : You don't answer the question - the 'field' which is asked for is cooking. You gave an explanation, but not the answer. $\endgroup$ – Tim Couwelier Apr 13 '15 at 11:06
  • 11
    $\begingroup$ Given that US measurements are completely ridiculous, I'm willing to believe this equation. Adopt the metric system, already! $\endgroup$ – Ian MacDonald Apr 13 '15 at 11:22
  • $\begingroup$ @TimCouwelier. I agree. I updated the answer accordingly. $\endgroup$ – Florian F Apr 13 '15 at 14:26
  • 1
    $\begingroup$ Hmm... we could use a unit of measurement based originally on an incorrect estimation of the earth's circumference and later an arbitrary length of a piece of platinum, or an semi-arbitrary distance that happens to be (within 2%) based on the distance light travels in a vacuum in a nanosecond. $\endgroup$ – Foon Apr 14 '15 at 14:07

Mathematically, I think the answer is


where this denotes

the 2-adic completion, i.e. the completion w.r.t. the 2-adic norm $||.||_2$,


the number field $\mathbb{Q}(\pi)/(\pi^2-2)$ - which is isomorphic to $\mathbb{Q}(\sqrt{2})$ as a number field, but has $\pi$ identified with $\sqrt{2}$.

This is a complete field (the question asks "in what field does..." and "I am looking for a complete answer") in which $\pi^2=2$.

My first idea was simply



we use the Greek letter $\pi$ to denote the number $\sqrt{2}$.

This isn't as silly as it looks, since the notation $\pi$ is often used in algebraic number theory for elements of $p$-adic completions of number fields rather than for $3.14159265358979...$. But I amended it as suggested by @Meelo since I think it makes slightly more sense with the actual number $\pi=3.14159265358979...$ identified with $\sqrt{2}$ via quotienting.

  • 8
    $\begingroup$ But Joel specified that he meant $\pi$ as the ratio of circumference/diameter in a circle. $\endgroup$ – Florian F Apr 13 '15 at 10:52
  • $\begingroup$ @FlorianF Hmm, good point. I'll have to think of some way of explaining that, maybe using a nonstandard definition of "circle" in a number field. $\endgroup$ – Rand al'Thor Apr 13 '15 at 11:02
  • $\begingroup$ Consider a circle constructed by drawing straight lines among four points...... ;) $\endgroup$ – Ian MacDonald Apr 13 '15 at 22:05
  • $\begingroup$ $\mathbb Q[\pi]/(\pi^2-2)$ is probably more correct. It's isomorphic, as a field, but more emphasizes the $\pi$. (Of course, one might ask: is it cheating to quotient out the relation you want?) $\endgroup$ – Milo Brandt Apr 13 '15 at 23:06
  • $\begingroup$ @Meelo Thanks - good idea! I've updated my answer accordingly. $\endgroup$ – Rand al'Thor Apr 13 '15 at 23:18

Here's an answer that goes in and out of figuration, becoming more figurative than literal and then more literal than figurative, stimulated by the hint.

The answer we arrrive at is that

the field is ancient Rome, or anywhere else that people have used Roman numerals.

The working is as follows.

We want $\pi^2$, i.e.

the result of taking $\pi$, the ratio between the circumference and diameter of a circle, and getting it to operate on itself to give us a square. Alternatively, to square something is to make a square out of it.

OK so

let's get figurative. How do we make a square out of the usual figure for $\pi$? Easy. Start by making a copy and turning it round:




stick the rotated copy onto the bottom:

We get


We have

now made a square. It's got bits sticking out of it (outside of the box), but we've still made a square by getting $\pi$ to operate on itself. So we've squared $\pi$.


Go back to being literal. What have we got? The Roman numeral for the number 2. So we've squared $\pi$ and got 2.


I realise the usage of literal here is questionable. Another weakness is that the meaning of $\pi$, which the setter stresses, doesn't get a look-in. Nonetheless, the train of thought goes from meaning to figure to figure to meaning, which fits nicely with the hint and works as a way of getting $\pi^2$ to equal 2.


I like commenter Lopsy's suggestion of this enter image description here

which seems to be a circle in Taxicab geometry AKA $L^1$ space. In this case however, $\pi = 2^2$, not $\pi^2 = 2$. It's possible that the OP made a mistake.

(Lopsy, you can see that $\pi=4$ not $2$ or $2\sqrt{2}$ because the length of each diagonal side is equal to $r+r$)

I tried to find a value of $p$ that made $\pi = \sqrt{2}$ in $L^p$ space but found $\pi$ was minimized at $p=2$ with the usual value of $\approx 3.1416$. It is equal to $4$ at $p=1$ and $p=\infty$ and diverges to $\infty$ as $p \to 0$. I don't think a circle is well-defined for $p \leq 0$.

  • 2
    $\begingroup$ I don't think that answering with the hope of a mistake in the OP is the correct way of solving a puzzle... $\endgroup$ – leoll2 Apr 14 '15 at 18:45
  • $\begingroup$ Haha I agree it's a long shot. $\endgroup$ – Hugh Allen Apr 15 '15 at 0:34

The Zeta function ζ(2) = π^2 $ζ(2) = π^2/6$ but then you want to know the field, the Riemann zeta function is used in quantum theory so that's your field.


stupid me, I went to look for a reference to Wikipedia for the answer and found out I was almost right:
enter image description here
I forgot i had to divide it by 6.

BONUS (unrelated to question):

the Quantum theory often makes odd statements which turn out to be true for example: $1 + 2 + 3 + 4 + 5 + .... = -1/12$ (so counting up all integers to infinity equals minus 1/12th) Link for explanation
As enforced by the police I have to add this is only true when used in several techniques called analytic continuation and Ramanujan sums. So stay in school kids, else the police will find you and they will correct you!

  • 15
    $\begingroup$ I am a member of the math police, whenever someone says that 1 + 2 + 3 + 4 + 5 + .... = -1/12 and links the shitty numberphile video I am contractually obliged to say this. "The series 1+2+3+4+... does not equal -1/12. It goes to infinity, as you would expect. BUT there are super interesting techniques, called analytic continuation and Ramanujan sums, which let you assign the value -1/12 to the series 1+2+3+4+..., and this is even useful in some crazy situations. But the series does not equal -1/12, in any ordinary sense of the word equal." $\endgroup$ – Lopsy Apr 13 '15 at 14:55
  • 1
    $\begingroup$ Sorry you are right but it's always such a great party trick :P sometimes one just has to bend the rules and portray something in a way that will make it just that bit more interesting :p $\endgroup$ – Vincent Apr 13 '15 at 14:59
  • $\begingroup$ Yup, I totally understand. It's a super cool fact! I've just seen a few people who hear it and then get confused when they actually learn about series, or worse, take it as evidence that math is some kind of cult that they'll never understand :( $\endgroup$ – Lopsy Apr 13 '15 at 15:03
  • 1
    $\begingroup$ Added a disclaimer, also shitty numberphile video? c'mon numberphile is fun, thanks to them i know math can be fun, sometimes. $\endgroup$ – Vincent Apr 13 '15 at 15:11
  • 1
    $\begingroup$ @Lopsy: If I was the math police, I wouldn't say the series goes to infinity. I might say that the sum of the series is infinity, or that the sequence of partial sums converges to infinity, but if I was being very precise, I would clarify that "sum" means to apply the operation you learn in calc II. (especially in a context where other summation operators are of interest) $\endgroup$ – Hurkyl Apr 13 '15 at 17:11

The answer is Cooking, where the pie has a piece cut out...

Not a full circle, so $\pi^2$ = 2

  • $\begingroup$ Can you expand a little on why you think this answer matches the question? I don't understand your answer as it is now. $\endgroup$ – Rand al'Thor Apr 14 '15 at 20:18
  • $\begingroup$ Mostly being silly/thinking outside the box. π is the circumference of a circle. If the circle is an actual pie, and you took a few slices out, it wouldn't be a full circle, or a full pie, so not a full π either. $\endgroup$ – AndyD273 Apr 14 '15 at 22:19
  • $\begingroup$ @AndyD273 $\pi$ is half a circle. Or more precisely an angle of $180^{\circ}$ or half a circumference of a circle with radius 1. Good guess though. Whenever I see $\pi$ I think of pie, but couldn't work out why squaring one would give you 2. $\endgroup$ – Bob Apr 14 '15 at 23:38

Likely wrong, but I wanted to add a different angle to the question, i.e. move a bit further outside the box.

Could be the touch-field of a pocket-calculator or other technichal device where touching the "Pi" key twice gives you 2 ? (Haven't found an according calculator though, yet.)


In physics, we sometimes use $\pi$ to mean the permutation operator that swaps two particles. If you swap two particles and then swap them again, you get back to the same state as before (...usually). Therefore, $\pi^2 = 1$.

  • 1
    $\begingroup$ OP definitely stated: "where is the ratio of the circumference of a circle to its diameter?" though your answer is quite interesting it's not a correct answer to the question. $\endgroup$ – Vincent Apr 14 '15 at 6:43

The simplest answer would seem to be "boxing", if we consider a ring to be synonymous with a circle.

If the diameter of the square is its diagonal and the circumference is its perimeter then the square of their ratio is 2.

  • $\begingroup$ Following the puzzling.stackexchange tradition, can you please add some details to your answer. $\endgroup$ – Mohit Jain Apr 14 '15 at 9:17
  • $\begingroup$ If the diameter of the square is its diagonal and the circumference is its perimeter then the square of their ratio is 2. $\endgroup$ – Alchymist Apr 14 '15 at 9:19
  • 2
    $\begingroup$ ... which is, of course, nonsense. It's 8. $\endgroup$ – Alchymist Apr 14 '15 at 9:26
  • $\begingroup$ @Alchymist So a MMA ring? $\endgroup$ – AndyD273 Apr 14 '15 at 22:20

In a strong gravitational field.

Or any other similar non-euclidean geometry that is sufficiently warped.

This video explains it with a nice visual demonstration using strechy fabric.

Imagine a circular trampoline with a weight at the centre. The edge of the trampoline remains fixed but as the weight increases the surface becomes more curved. The distance in straight line following the trampolines surface from the edge to the centre (its radius) becomes greater. When it is finally curved just the right amount the the ratio of the circumference to the diameter will reach $\sqrt2$

  • $\begingroup$ Would you mind explaining a bit more how this would make $\pi^2 = 2$? $\endgroup$ – Tryth Apr 15 '15 at 8:39
  • $\begingroup$ I have to admit that my practical knowledge of geometry is very much euclidean so I can't calculate for you how much 2d space needs to be warped to reach $\pi^2=2$ $\endgroup$ – Bob Apr 15 '15 at 9:36

protected by Aza Apr 17 '15 at 19:03

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.