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I want to know if it is possible to find two arithmetic progressions of three square numbers, with the same common difference:

\begin{align} \ & a^2 +r = b^2 \\ & b^2 +r = c^2 \\ & a^2 +c^2 = 2\,b^2 \\ \end{align}

and

\begin{align} \ & d^2 +r = e^2 \\ & e^2 +r = f^2 \\ & d^2 +f^2 = 2e^2 \\ \end{align}

where $a,b,c,d,r \in \Bbb N$.

Here is an example that almost works:

\begin{align} \ & 23^2 +41496 = 205^2 \\ & 205^2 + 41496 = 289^2 \\ & 23^2 +289^2 = 2\,(205)^2 \\ \end{align}

and

\begin{align} \ & 373^2 + 41496 = 425^2 \\ & 425^2 + 41496 = \color{#C00000}{222121} \\ & 23^2 + \color{#C00000}{222121} = 2\,(205)^2 \\ \end{align}

where the difference is $41496$, but the last element isn't a square number.

I can't find an example of two progressions with three numbers and the same common difference. Could you demonstrate that such progressions are nonexistent using reductio ad absurdum to this statement?

Théophile
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Juan Alfaro
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  • The correct formulation of the problem - is half of its solution. 2 and 3 equations are equivalent to each other. It is necessary to solve a system of 2 equations. As elementary as it is solved with a different view of the difference of squares. – individ Oct 13 '17 at 16:19
  • $$r=b^2-a^2=c^2-b^2$$ It is necessary to solve this equation. More than anything do not have to deal. – individ Oct 13 '17 at 16:23
  • https://artofproblemsolving.com/community/c3046h1051335_pythagorean_triple – individ Oct 13 '17 at 16:38

2 Answers2

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$$(a,b,c,d,e,f,r)=(1,29,41,23,37,47,840)$$ satisfies $$a^2 +r = b^2,\quad b^2 +r = c^2,\quad a^2 +c^2 = 2b^2$$ $$d^2 +r = e^2,\quad e^2 +r = f^2,\quad d^2 +f^2 = 2e^2$$

mathlove
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There are infinitely many solutions to the system,

\begin{align} \ & a^2 +r_1 = b^2 \\ & b^2 +r_1 = c^2 \\ & a^2 +c^2 = 2b^2 \\ \hline \ & d^2 +r_2 = e^2 \\ & e^2 +r_2 = f^2 \\ & d^2 +f^2 = 2e^2 \\ \end{align}

with $\color{blue}{r_1=r_2}$. Eliminating $r_1$ between the first two equations (and similarly for $r_2$), one must solve the Pythagorean-like,

$$a^2+c^2=2b^2\\ d^2+f^2=2e^2$$

which has solution,

$$a,b,c = p^2 - 2q^2,\; p^2 + 2p q + 2q^2,\; p^2 + 4p q + 2q^2\\ d,e,f = r^2 - 2s^2,\; r^2 + 2r s + 2s^2,\; r^2 + 4r s + 2s^2$$

Hence,

$$r_1 = -a^2+b^2 = -b^2+c^2 = 4 p q (p + q) (p + 2 q)\\ r_2 = -d^2+e^2 = -e^2+f^2 = 4 r s (r + s) (r + 2 s)$$

Thus one must solve,

$$p q (p + q) (p + 2 q) = r s (r + s) (r + 2 s)$$

This is essentially the same equation in this post, hence one solution (among many) is,

$$p,\;q = 2 n (m + 6 n),\; m (m + 4 n)\\ \;r,\;s = m (m + 2 n),\; 4 n (m + 3 n)$$

For example, let $m,n = 1,1$, then,

$$a,b,c = 146, 386, 526\\ d,e,f = 503, 617, 713\\ r_1=r_2= 127680$$

and so on.

Tito Piezas III
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  • In my opinion, this answer is far better than the accepted answer. Well done! $(+1)$ – Mr Pie Jul 10 '19 at 10:52
  • @MrPie: Thank you, MrPie. I was late to the party, unfortunately. – Tito Piezas III Jul 10 '19 at 10:53
  • I think you deserve a tick/checkmark, but unfortunately, there can only be one... fortunately however, giving your answer some time for further review, I might give you a bounty to make up for the tick. Many answers are in the same boat as you, but you are nearing 30k rep, so... special occasion ;) – Mr Pie Jul 10 '19 at 10:56
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    @MrPie: That's ok, I enjoy these type of questions. But that's a nice thought. :) – Tito Piezas III Jul 10 '19 at 10:58