Squares in Progressions

Find all arithmetic progressions {x, y, z} where {x, y, z} are all squares.

So we need to find integers {a,b,c} such that {b^2-a^2 = c^2-b^2,} ie., {a^2+c^2=2b^2.} {\left(\frac{a}{b}\right)^{2}+\left(\frac{c}{b}\right)^{2}=2.} Hence we get a rational point on the circle {X^2+Y^2=2.} {(X,Y)=(1,1)} is a point on the circle and now using a pencil of lines passing though {(1,1)} parametrized by slope {m}, we see that all the point are given by

\displaystyle X=\frac{m^{2}-2 m-1}{m^{2}+1}, \quad Y=\frac{m^{2}+2 m-1}{m^{2}+1}
\displaystyle m=\frac{1}{2} \implies X=\frac{-7}{ 5} ,Y=\frac{1}{ 5} \implies (a,b,c)=(1,25,49)
\displaystyle m=\frac{-1}{3} \implies X=\frac{-1}{ 5} ,Y=\frac{7}{ 5} \implies (a,b,c)=(1,25,49)
\displaystyle m=\frac{3}{4} \implies X=\frac{-31}{ 25} ,Y=\frac{-17}{ 25} \implies (a,b,c)=(289,625,961)
\displaystyle m=\frac{-5}{3} \implies X=\frac{23}{ 17} ,Y=\frac{-7}{ 17} \implies (a,b,c)=(49,289,529)
\displaystyle m =\frac{7}{5} \implies X=\frac{23}{ 17} ,Y=\frac{-7}{ 17} \implies (a,b,c)=(49,289,529)

What if you fix the common difference to be {n}?

Congruent Numbers, Elliptic Curves:

Find rational triples {(a,b,c)} such that {b^2-a^2=c^2-b^2=n.}
\displaystyle (a,b,c) ~|~ ~b^2-a^2=c^2-b^2=n \iff (k, m)~ |~~ nk^2=m^2-m
where
\displaystyle (a,b,c) \mapsto m =\frac{c-b}{a-b}, ~k=\frac{m^{2}+1}{2 b}=\frac{2 b-a-c}{(a-b)^{2}}
\displaystyle (m,k) \mapsto \left(a=\frac{m^{2}-2 m-1}{2 k}, b=\frac{m^{2}+1}{2 k}, c=\frac{-m^{2}-2 m+1}{2 k}\right).

So we have the map

\displaystyle (a,b,c) \mapsto nY^2=X^3-X
But
\displaystyle nY^2=X^3-X \mapsto Y^2=X^3-n^2X
\displaystyle (X, Y) \mapsto\left(nX, n^2Y\right)

Numbers {n} which have rational solutions on this elliptic curve with {Y\neq 0} are called congruent numbers. By the above, they exactly correspond to arithmetic progressions in squares with common difference {n.}
{(a,b,c) \mapsto (A= c+a,B=c-a, C=2b)} gives a map to right triangle {(A,B,C)| A^2+B^2=C^2}, who area { \frac{AB}{2}=n.}

Fact: The only points of finite order on {Y^2=X^3-n^2X} are {(0,0),(n, 0),} and {(-n, 0).}

Hence we can use the group law on the curve to generate infinitely many rational points on this curve. And therefore if there is one arithmetic progression we can find infinitely many arithmetic progressions in squares with common difference {n.}

{(1,25,49)} corresponds {n=24} and {(X,Y)=(-12,72)} on {Y^2=X^3-24^2X.}

Computing {2 P=(25,-35), \quad 3 P=\left(-\frac{6348}{1369},-\frac{2568456}{50653}\right), \quad 4 P=\left(\frac{1442401}{4900}, \frac{1726556399}{343000}\right)}

we get the progressions

\displaystyle \left(\frac{1151}{70}, \frac{-1201}{70}, \frac{1249}{70}\right), \left(\frac{4319999}{1319901}, \frac{-7776485}{1319901}, \frac{-10113607}{1319901}\right)
\displaystyle \left(\frac{1727438169601}{241717895860}, \frac{2094350404801}{241717895860}, \frac{-2405943600001}{241717895860}\right)

Open: How do we classify the congruent numbers {n} ? And what about the rational solutions (what is the rank, generators of the rational points — what are all progressions in squares?

For instance, the following family of curves have rank at least 2 for almost all {t.}

\displaystyle y^{2}=x\left(x^{2}-f(t)^{2}\right) , f(t) =t(t+1)(3 t+1)\left(9 t^{4}+24 t^{3}+26 t^{2}+8 t+1\right)

INTEGERS: (Congruum Problem)
The above discussion is about rational progressions. What if we ask for the progressions to be in integers? That is we need integer solutions for

\displaystyle \begin{aligned} &x^{2}+n=a^{2}\\ &x^{2}-n=b^{2}. \end{aligned}

Fibonacci: All the integer solutions to these equations are given by

\displaystyle \begin{aligned} &x=p^{2}+q^{2}\\ &n=4 pq\left(p^{2}-q^{2}\right) \end{aligned}

Proof Sketch: Observe that
\displaystyle a^2+b^2=c^2 \iff (a+b)^2+ (a-b)^2 =2c^2
hence above parametrization of the solutions comes from parametrization of Pythagorean triples.

FOUR TERMS?

What about progressions of length {4}?
Fermat’s four squares theorem: There are no four distinct rational squares in arithmetic progression.


We can suppose the the progression looks like {x-6k, x-2k, x+2k, x+6k}, all of them squares. {n=4k} is the common difference. So we have an integer point on

\displaystyle y^{2}=\left(x^{2}-4 n^{2}\right)\left(x^{2}-36 n^{2}\right)=\left(x^{2}-20 n^{2}\right)^{2}-256 n^{4}
\displaystyle \left(16 n^{2}, y, x^{2}-20 n^{2}\right) is a Pythagorean triple.
So we have for some {u,v}
\displaystyle 16 n^{2}=4uv, x^{2}-20 n^{2}=4 u^{2}+v^{2} which implies
\displaystyle u=4 A^{2} \text { and } v=D^{2}
\displaystyle 4 u^{2}+v^{2}=x^{2}-5 u v
\displaystyle 4 u^{2}+v^{2}+5uv= (4u+v)(u+v)=x^{2}
{4 u+v=16 A^{2}+D^{2}} and {u+v=4 A^{2}+D^{2}} have to be squares.

Again using Pythagoren parametrization we get {2 U V=2 A} and {U^{2}-V^{2}=\pm D} and {4 U^{\prime} V^{\prime}=4 A} and {4 U^{\prime 2}-V^{\prime 2}=\pm D}

{U V=U^{\prime} V^{\prime}} implies that there are integers {2 a, b, c, d} so that {\pm D=4 a^{2} b^{2}-c^{2} d^{2}=16 a^{2} c^{2}-b^{2} d^{2}.} {b^{2}\left(4 a^{2}+d^{2}\right)=c^{2}\left(16 a^{2}+d^{2}\right)} gives us that {4 a^{2}+d^{2}} and {16 a^{2}+d^{2}} are squares. Using the above argument, in reverse gets you {4} squares in progression with common difference { 4ad} which is smaller than {AD,} the common difference we started with.
By descent by are done!

In the background of the above elementary argument is the curve{E: y^{2}=x(x+1)(x+4)}– the argument amount to showing that there no rational points which corresponds to 4 term progressions in squares.(We need points of infinite order on this curve which don’t exist.)

https://kconrad.math.uconn.edu/blurbs/ugradnumthy/4squarearithprog.pdf
https://kconrad.math.uconn.edu/blurbs/ugradnumthy/3squarearithprog.pdf
https://arxiv.org/pdf/0712.3850.pdf

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