Arithmetic progression of four square number with the same common difference

Asked Oct 13 '17 at 08:23

Active Oct 13 '17 at 08:38

Viewed 77 times

I want to know, if is possible found the arithmetic progression of four square number, with the same common difference.

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

where a,b,c,d,r $\in\mathbb{N}$

Hence, I found only triple succession.

For example: 1,5,7 because \begin{align} \ & 1^2 +24 = 5^2 \\ & 5^2 +24 = 7^2 \\ \end{align} where the ratio is 24.

but I can't found a example with four number and the same common difference. at worst. Could you demonstrate that nonexistent using reductio ad absurdum to this statement?.

edited Oct 13 '17 at 08:38

asked Oct 13 '17 at 08:23

Juan Alfaro

It's usually called a common difference, not a ratio. – Steven Alexis Gregory Oct 13 '17 at 08:34
1

Ok, but I'm just learning English. Thanks for your comment – Juan Alfaro Oct 13 '17 at 08:37

Arithmetic progression of four square number with the same common difference

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