This is a great question.

For an answer to this interesting question, see

arXiv:1611.07441v1

at

http://arxiv.org/archive/math

For periodic variants of the square peg problem, see Section 4, starting on page 12.

For the algebraic square peg problem in a 2014 M.Sc. thesis, see

arXiv:1403.5979v1

also

at 

http://arxiv.org/archive/math

@James F Peters · Thanks!  

The paper (arXiv:1611.07441v1) by Prof. T. Tao is  also found at his research blog website cited above.  Please visit that website (https://terrytao.wordpress.com/2016/11/22/an-integration-approach-to-the-toeplitz-square-peg-problem/#comment-474751) for more details about his paper and other great stuff (mathematics).

@David Cole: ... and other great stuff (mathematics).

Many thanks for your comments and very helpful observations.   We agree: the work bhy Prof. T. Tao leads to great stuff in mathematics.

I believe it might be possible to construct a closed curve that has no Toeplitz square. Consider the map of Africa, but with smoothed edges and elastic boundaries. For every candidate of Toeplitz square I can find, there is a possible stretching of some boundary segment that defeats the candidate without creating another candidate. Proof is hard, but if possible would be a mathematical description of the flexible map example.

In general an unsymmetrical pear shape might prevent a Toeplitz square if the large end is wider in the transverse than it is in the axis, and the boundaries are a bit flexible.

Maybe someone will find a way to disprove this. It's a hard proof because of the many degrees of freedom in making the curve and the strict requirement of 4 equal sides and 4 right angles, leaving only translation, rotation, and length of a side as freedoms for the Toeplitz square. In mathematical terms the number of unknowns can increase faster than the number of equations. 

Notice in the asymmetrical pear if the large end is narrower in the transverse than it is in the axis, there is usually a Toeplitz square for it. 

@Jerry Decker  Very interesting!  But I believe there's something beautifully invariant about geometry and topology that our current problem seems to capture quite elegantly.  

We inscribe a general Jordan curve in a square, and that Jordan curve inscribes a square!   Wow!   That's too beautiful -- it must be true!

In formulating a proof of the Inscribed Square Conjecture,  we can  distinguish between Jordan curves which obey our assumption (simple closed curves which does not have four points on it that form an inscribed square of any size from the unit square, s1,  to sn less than the size of the square which inscribes it) and those which do not obey our assumption.

A Jordan curve which obeys our assumption is exceptional,  and we can label it, Je. And a Jordan curve which  does not obey our assumption is regular,  and we can label it, Jr.

Our task is to prove that Je does not exist.  And we can use Jr to help us prove that result.  How?  Hmm...

Reference:   Jerry Decker's post (above) is quite interesting, and helpful. 

Can we develop Je from Jr?

Inscribed Square Conjecture is true!  The mapping of all  four points of the general Jordan curve will eventually converge to  the four corners or vertices, respectively, of a square of size, s1,  or greater.

"Hmm.   Where we begin is where we should end -- back to square one."  

Formulation of Proof:

Key Ideas: Orientation of points on Je  about center point (Cp),   Invariant properties of Je,   minimum unit square (s1),  Convergence via limiting process of farthest two points on Je.

Let P1, P2, P3, P4  be four distinct points on Je such that the line segments, P1P3  and  P2P4,  are parallel.  And likewise, the line segments, P1P2  and P3P4, are parallel. Furthermore,  the line segments, P1P3 and P2P4, may or may not be perpendicular to the line segments, P1P2 and P3P4.  But the lengths of  the four line segments are unequal when line segments, P1P3 and P2P4, are perpendicular to  the line segments, P1P2 and P3P4, according to our prime assumption.  

The four points are  relative to our center point, Cp, which is the intersection of the two diagonals of S which is the square which encloses Je.  We construct an arc angle, ρ, about Cp that measures the angular arc of the area bounded inside by our four line segments.  We assume line segment, P1P2, is nearest to Cp and opposite to Cp,  and the line segment, P3P4, is farther away from Cp and opposite to Cp.   And the arc angle, ρ12, measures the angular arc between points, P1 and P2.  The arc angle, ρ34, measures the angular arc between points, P3 and P4.

Our square, S, has vertices, P1 = (0, 0), P2 = (0, 10),  P3 = (10,  0), and P4 = (10, 10).  And  two farthest points on Je are the points, of S.  And according to the properties of Je as defined in a previous post, there could be a point of Je  at  either,  P2  or  P3  of  S, but not at both points.

Let's assign the next two farthest points on  Je less than the length of line segment, P1P4, to the vertices, respectively,  P12 and P42,  of the square, sn, which is less than the size of S.  And according to the properties of Je, there could be a point of Je at either of the two vertices, P22 or P32 of sn, but not at both points.

Let's assign the next two farthest points on Je less than the length of line segment, P12P42, to the vertices, respectively, P13 and P43, of the square, sn-1, which is less than the size of sn. And according to the properties of Je, there could be a point of Je at either of the two vertices, P23 or P33 of sn-1, but not at both points.

However, if we continue that process of assigning the next two farthest points on Je less than the length of line segment, P1mP4m,

for some m > 2, to the vertices, respectively, P1(m+1) and P4(m+1), of the square, sn-m, which is less than the size of the previous square, sm-n+1,  we discover there exists a minimum unit square, s1 = sm-n,  for which there are two more points of Je at vertices,  respectively, P2(m+1) and  P3(m+1) of s1.  

...

Remarks:  This proof is not complete!

Q1:  How does the boundary of S impacts Je?

Q2:  How does J'e = S - Je impacts Je?

Q3:  How do  curvature, smoothness,  simple closed curve,  and connectedness  of  points along Je affect the prime assumption?

Q4:  How does one make use of normals and tangents to help solve our problem?

Q5:  Can we develop Je from  from parts  of Jr where there are no inscribed squares?

Q6:  Can computer simulations help verify the Inscribed Square Conjecture?

Q7:  As we move along Je in four different places, pointwise,  what criteria can we develop to determine effectively which parts of Je  may have inscribed squares?

Q8:  If we begin with a square inscribed in a circle which is a Jordan curve,  what are precisely the transformation group operations, contraction, translation and rotation, etc.,  which generates a different regular Jordan curve, Jr, or an exceptional Jordan curve, Je, with or without inscribe squares, respectively?

The more I meditate on a solution to the Inscribed Square Problem (ISP),  the more I realize ISP is a very elegant problem which requires a very elegant solution. It may take me six months of study and research to crack it.   Meanwhile, I am in awe of this wonderful problem for several reasons...

FORMULATION OF A SOLUTION TO THE INSCRIBED SQUARE PROBLEM:

Key Ideas Part 1:  Properties of the exceptional Jordan curve, Je, 2π Rotation of the two Diagonals of the enclosing square, S, of Je,  about the Center Point, Cp = (0,0) that intersect at points along Je and the 2π Rotation and translating of  perpendicular line segments intersecting Je about the two rotating Diagonals of S. And the use of convergence to help solve the Inscribed Square Problem.

First Key Equation Formulation:

Let P1, P3, P5, and P7 be distinct points on our exceptional Jordan curve, Je∈ R2 (see the definition of Je in a previous post here) such that the following equation involving Euclidean norms is true.

1.  ||P1 - P5||2 = ||P1 - P3||2  +  ||P3 - P5||2  =  ||P1 - P7||2  +  ||P7 - P5||2  where

0  0, Δ2 > 0, Δ3 > 0,  and Δ4 > 0 such that:

2.  (||P1 - P5|| ±  ||Δ ||)2 = (||P1 - P3|| ± ||Δ1||)2  +  (||P3 - P5|| ∓ ||Δ2||)2 

=  (||P1 - P7|| ± ||Δ3||)2  +  (||P7 - P5||  ∓  ||Δ4||)2 

such that 0 <  ||P1 - P5|| ± ||Δ|| ≤ dmax...

Note:  The important symbols, ± and ∓,  requires further important detailed derivations here. 

3.  Je  is enclosed in a square, S,  whose diagonal length equals the greatest distance between two distinct points, P1 and P5on Je.  So the vertices of S are P1 = (-dmax  /  √2, -dmax  / √2), P30 = (-dmax  /  √2, dmax  /  √2),

P5 = (dmax  /  √2, dmax  /  √2), and P70 =(dmax/ √2, -dmax / √2).  And the Center Point, Cp =(0,0), which is the intersection point of the two diagonals, P1P5  and P30P70,  of S. 

Note:  Only  one  of two points, P30 and P70, may belong to Je.

4.  Our exceptional Jordan curve, Je, is a general piecewise mixed Cm simple closed curve of the xy-plane where the integer, m, is greater than or equal to one.

We define Je enclosed  by S  to be

y(x) = {

y0(x),  x0 ≤ x ≤ x1; 

y1(x),  x1 ≤ x ≤ x2;

y2(x),  x2 ≤ x ≤ x3;

y3(x),  x3 ≤ x ≤ x4;

...

yn(x),  xn ≤ x ≤ x0;

where y0(x0) = yn(x0), y0(x1) = y1(x1), y1(x2) = y2(x2), y2(x3) = y3(x3), ...,

yn-1(xn) = yn(xn), otherwise, 

y0(x), y1(x), y2(x), y3(x), ..., yn-1(x), and yn(x) are strictly unequal functions of x according to the definition of Jordan curve.

Note:  The integer, n, is greater than or equal to two.

In addition, we have the following system of differential equations:

dy0(x0)/dx  =  dyn(x0)/dx; 

dy0(x1)/dx  =  dy1(x1)/dx;

dy1(x2)/dx  =  dy2(x2)/dx;

dy2(x3)/dx  =   dy3(x3)/dx;

...

dyn-1(xn)/dx  =  dyn(xn)/dx.

5.  According to our definition of Je, the following conditions (I, II, III, and IV) must not be satisfied together for all x such that

-dmax  / √2  ≤   x  ≤  dmax / √2  and

0 < dmin ≤  || ( x , yk(x) ) - ( x ± a, yl(x ± a) ) || =   2d  ≤  dmax with a > 0 and

b > 0:

I.    || ( x , yk(x) ) - ( x ± b, yj(x ± b) ) ||  =  d * √2  and

II.   || ( x , yk(x) ) - ( x ∓ b, yq(x ∓ b) ) ||  =  d * √2  and

III.   || ( x ± a , yl(x ± a) ) - ( x ± b, yj(x ± b) ) || =  d * √2,  and

IV.  || ( x ± a , yl(x ± a) ) - ( x ∓ b, yq(x ∓ b) ) ||  = d * √2 

where  j, k, l, and q are positive integers between one and the integer,

n ≥ 2, inclusively.

Note: Our work (...) is almost done!  :-)

Can we disprove our assumption about Je (see 5 above) by means of analytic geometry (tangents and normals of curves, etc.), topology, insight, ingenuity, and convergence?

Key Ideas Part 2:  Parallel Tangents of corresponding curves of y(x) and Parallel Normals of corresponding curves of y(x) according to the properties of simple closed curve, Je, Winding Numbers, and the Convergence to an Inscribed Square of Je.

Hmm...  While considering the properties of Je and the properties of y(x), we need to determine if the following  system of linear functions or linear equations is always true for some four distinct points on Je: 

w1(x) = r * x + t1 (a line segment through two distinct points on Je, P1 and P3) ;

w2(x) = r * x + t2 (a line segment through two distinct points on Je, P5 and P7);

w3(x) = (-1 / r) * x + t3 (a line segment through two distinct points on Je, P1 and P7);

w4(x) = (-1 / r )* x +  t4 (a line segment through two distinct points on Je, P3 and P5) such that  0 < | t1 - t2 | = | t3 - t4 | <  dmax.

Hmm...  Moreover, we are also mindful of inscribed circles of Je that intersect Je at least four distinct points.  And we shall seek out appropriate inscribed squares of these inscribed circles of Je or inscribed squares adjacent to them.  

Note I:  There may be one curve, yk(x), of Je which  goes through one vertex of an inscribed square or connects two vertices of an inscribed square, or there  may be two distinct curves, yk(x) and yl(x), of Je that join at a common vertex of an inscribed square of Je.  

And possible configurations  of two or more concave or convex curves of Je and their corresponding orientations between subtended angles (https://en.wikipedia.org/wiki/Subtended_angle) must be thoroughly investigated while being mindful how Square, S, impacts those curves.

Note II:  If we begin with a square inscribed in a circle which is a Jordan curve,  what are precisely the transformation group operations, contraction, translation and rotation, etc.,  which generates a different regular Jordan curve, Jr, or an exceptional Jordan curve, Je, with or without inscribed squares, respectively?

Note III:  We believe we have sufficient information and knowledge to complete our proof of the Inscribed Square Conjecture.

The Key and General Idea for a proof of the Inscribed Square Conjecture:

The Square, S, which inscribes the Jordan curve, Je,  induces inscribed squares  in  Je  by  design (Generally,  increasing the number, n, of curves of Je  which are yk(x), where 2 ≤ k ≤ n, implies more winding curves of Je,  more closeness between curves of Je, and therefore, more inscribed squares in Je).   

That statement contradicts Je  which must not have any inscribed squares.

PROOF OF THE INSCRIBED SQUARE CONJECTURE (ISC):

Let  sn (please see a previous post here for more information) be largest square less than the size of Square,  S,  which is not inscribed by Je such that

0 < s1 < s2 < s3 < ... < sn-1 < sn < S where sk, 1 < k ≤ n, are squares which are also not inscribed by Je.  Then the probability that Je does not inscribe any square, sk, is

Prob(Je does not inscribe any square, sk,  for  1 ≤  k  ≤ n )

= Prob(Je does not inscribe  square, s1 | Je does not inscribe square, s2 )

* Prob(Je does not inscribe square, s2 | Je does not inscribe square, s3 )

* Prob(Je does not inscribe square, s3 | Je does not inscribe square, s4 )

...

*  Prob(Je does not inscribe square, sn-1 | Je does not inscribe square, sn)

* Prob(Je does not inscribe square, sn )

= Prob(Je does not inscribe square, s1) 

since we assumed:

Prob(Je does not  inscribe any square, sk, for  1 < k  ≤ n )  = 1.

However,

Prob(Je does not inscribe square, s1)  =  (4 / 5)i  ⟶ 0 as i ⟶ ∞.

Notes: 

1.  The squares, s1, inside S, may have zero vertices intersecting Je inside S;

Or  2. The squares, s1, inside S, may have one vertex intersecting Je inside S;

Or  3. The squares, s1, inside S, may have two vertices intersecting Je inside S;

Or  4.  The squares, s1, inside S, may have three vertices intersecting Je inside S;

 Or 5. The squares, s1, inside S, may have four vertices intersecting Je inside S.

[Our probability space is the space of  appropriate sk in close neighborhood of Je.]

Therefore,  Prob(Je does not inscribe any square, sk, for 1 ≤ k ≤ n ) = 0  since:

Prob(Je does not inscribe square, s1) = 0.

Hence, the Inscribed Square Conjecture (ISC) is true!

*****

Final Notes:  The transformation of Je as it tries to avoid generating any inscribed squares, sk,  for  1 ≤ k ≤ n is quite fascinating, and should be studied thoroughly.  

Moreover, since a Jordan Curve is topologically equivalent to either a circle or a square, can we formulate a solution to the Inscribed Square Problem for any given Jordan Curve as a solution to a specific minimum optimization problem?  We observe that the perimeter of a circle is greater the perimeter of an inscribed square whose vertices intersect that circle and a Jordan Curve...

And of course, our assumption, 

Prob(Je does not inscribe any square, sk, for 1 < k ≤ n ) = 1,  

can be challenged immediately based on the possibility that for some k between 2 and n, inclusively, there may exist at least one inscribed square, sk, in Je.

Our S can be generalized to various polygons which inscribes Je at several points ...  And the Generalized Inscribed Polygon Conjecture would be valid too under various stringent conditions (e.g. inscribing at most four vertices of regular polygons).

Of course, we can also add more technical details (please see reference idea below) to our proof of the inscribed square conjecture. Only time will tell.

Important Reference Idea From Previous Answer:

"The Square, S, which inscribes the Jordan curve, J_e,  induces inscribed squares  in  J_e  by  design."

Relevant Reference Resource:

'Differential Equations: Geometric Theory', by S. Lefschetz, 1977, ISBN: 0-486-63463-9.

David Cole nice work. I believe you could argue that for any square of side S the possible number of simple closed curves connecting the vertices is either finite or infinite. If infinite, then the conjecture is true because infinity includes all curves by scaling of S, but if the number of possible curves is finite then the conjecture is false because some curves are excluded. In my previous examples of flexible Africa and asymmetric pear I have taken the finite point of view by placing restrictions on ratios of distances.