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I was reading a pdf (Progressions of Squares by Tom C. Brown1, Allen R. Freedman2, and Peter Jau-Shyong Shiue2) (link: https://www.sfu.ca/~vjungic/tbrown/tom-9.pdf) and there the author on Page no. 1, Introduction paragraph 1 says: three-term arithmetic progressions occur in abundance among the squares: take any Pythagorean triple, a$^2$+b$^2$ = c$^2$; then (b-a) $^2$; c$^2$;(b + a) $^2$ is clearly a 3-term AP with common difference 2ab. It’s also easy to show that every 3-term AP of squares has this form.

I am trying to prove the last line of this statement but so far not able to proceed farther then this
2b$^2$ = a$^2$ + c$^2$ when a$^2$, b$2$, c$^2$ are in A.P and gcd(a,b,c) = 1.

Claim: The common difference of the given A.P. is even.

Proof: Since, all a, b, and c can't be even as gcd(a,b,c) = 1. Therefore, at least one of them is odd. Since, 2b$^2$ = a$^2$ + c$^2$. Therefore R.H.S. of this equation must be even as L.H.S. is even. Possible parity of (a,c) is either (even,even) or (odd,odd). If (a,c) = (even,even) and since one of them must be odd. Therefore, b is odd. But L.H.S is only divisible by 2 while R.H.S. is divisible by 4, which is not possible. Therefore, (a,c) = (odd,odd). And hence, all three a, b and c are odd. And there common difference is even

Bill Dubuque
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Raj Pratap Singh
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    What have you tried? Seems to be just a one line algebraic substitution. – Calvin Lin Oct 29 '24 at 22:35
  • I know that's where I am stuck. I thought using mod and could only conclude that both a and c must be odd. After that no idea has come to me. – Raj Pratap Singh Oct 29 '24 at 22:39
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    To get started: I suggest, prove that the period in any such three term progression must be even. – lulu Oct 29 '24 at 22:41
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    @RajPratapSingh $4,100,196$ is an arithmetic progression of squares with no odd numbers. – Henry Oct 29 '24 at 22:52
  • @Henry you are right. I should have mention that a, b and c are primitive. So, the example you have given is similar to 1,25,49. Sorry my bad for not mentioning this earlier. – Raj Pratap Singh Oct 29 '24 at 22:56
  • I don't understand. You proved the period is odd and from that you deduce that it is even? I suggest: edit your post to add all the assumptions you intended and to include the proof that the period must be even. – lulu Oct 29 '24 at 22:59
  • @lulu thanks for correcting me. I have edited my post. – Raj Pratap Singh Oct 29 '24 at 23:13
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    This question is similar to: Parametric characterization for $x^2 + y^2 = 2z^2$. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. Found using an Approach0 search. Note one thing not explicitly mentioned in the answer there is that, with your $a$ and $c$ ... – John Omielan Oct 29 '24 at 23:16
  • (cont.) having the same parity, then their sum and difference are both even, so an integers results when quantity is divided by $2$. Since $c\gt a$, have $e=\frac{c+a}{2}$ and $f=\frac{c-a}{2}$ be these $2$ positive integers. Then subtracting the second from the first gives $a=e-f$, while adding the $2$ gives $c=e+f$. Thus, your equation of $2b^2=a^2+c^2$ becomes $2b^2=(e-f)^2+(e+f)^2;\to;2b^2=(e^2-2ef+f^2)+(e^2+2ef+f^2);\to;b^2=e^2+f^2$, i.e., a Pythagorean triple. Finally, note this argument doesn't require $\gcd(a,b,c)=1$. – John Omielan Oct 29 '24 at 23:22
  • See also Fibonacci's Lost Theorem $\ \ $ – Bill Dubuque Oct 30 '24 at 00:24

2 Answers2

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Let $x^2,y^2,z^2$ three squares in arithmetic progression. Since $x^2+z^2=2y^2$ we have that $x$ and $z$ have the same parity.

Then the numbers $a=\frac{z-x}{2}$, $b=\frac{z+x}{2}$ and $c=y$ are integers and satisfy $a^2+b^2 = c^2$.

jjagmath
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We begin with Euclid's formula shown here as $$A=m^2-n^2\qquad B=2mn\qquad C=m^2+n^2\qquad m>n$$

We now take

\begin{align*} \bigg((m^2+n^2)-(m^2-n^2)\bigg)^2&=4n^4\\ (m^2+n^2)^2&=m^4 + 2 m^2 n^2 + n^4\\ \bigg((m^2+n^2)+(m^2-n^2)\bigg)^2&=4m^4 \end{align*}

and the differences are $\quad m^4 + 2 m^2 n^2 - 3 n^4 \quad\text{or} \quad n^4 + 2 m^2 n^2 + 3 m^4 \quad$ which do not match the pdf. However, if we use the parameters for generating these terms we get

$$ m^4 + 2 m^2 n^2 - 3 n^4 = 2^4 + 2(2^2)(1^2) - 3(1^2) = 21\\ n^4 + 2 m^2 n^2 + 3 m^4 = 1^4 + 2(2^2)(1^2) + 3(m^4) = 57 $$

And this does not make sense either. The only way we can show that the pdf statement is true is to use the actual triple values. With $\,(3,4,5),\,$ we get $\quad(4-3)^2=1\quad 5^2=25\quad (4+3)^2=49\quad$ and the differences are indeed $\,2(3)(4)=24.\quad$ If we use $\,(5,12,13),\,$ we get $\quad(12-5)^2=49\quad13^2=169\quad(12+5)^2=289\quad$ and the differences are $\,2(5)(12)=120.\quad$ This works for any triple we may try but I do not know how to prove it algebraically.

poetasis
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