3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – individ Oct 13 '17 at 16:19
  • $\begingroup$ $$r=b^2-a^2=c^2-b^2$$ It is necessary to solve this equation. More than anything do not have to deal. $\endgroup$ – individ Oct 13 '17 at 16:23
  • $\begingroup$ artofproblemsolving.com/community/… $\endgroup$ – individ Oct 13 '17 at 16:38
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In my opinion, this answer is far better than the accepted answer. Well done! $(+1)$ $\endgroup$ – Feeds Jul 10 at 10:52
  • $\begingroup$ @MrPie: Thank you, MrPie. I was late to the party, unfortunately. $\endgroup$ – Tito Piezas III Jul 10 at 10:53
  • $\begingroup$ 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 ;) $\endgroup$ – Feeds Jul 10 at 10:56
  • 1
    $\begingroup$ @MrPie: That's ok, I enjoy these type of questions. But that's a nice thought. :) $\endgroup$ – Tito Piezas III Jul 10 at 10:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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