Timeline for Special arithmetic progressions involving perfect squares
Current License: CC BY-SA 3.0
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| Feb 22, 2012 at 19:21 | comment | added | Michael Stoll |
To complete this: From $d \ge 4bc$, it follows easily that $a,b,c,d$ cannot be an AP. Note also the following: Acdording to Lemma 13 in Dujella's paper linked to in his answer above, if $a,b,c$ is a Diophantine triple with $a < b < c$, then $c = a + b + 2 \sqrt{ab+1}$ or $c \ge 4ab$. If $a,b,c$ form an AP, then the second possibility cannot occur, and the first (together with the AP condition) then implies that $b$ is even and $y^2 - 3(b/2)^2 = 1$, where $y = b-a = c-b$. So it is indeed the case that all such triples come from this Pell equation. (This gives another proof.)
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| Feb 12, 2012 at 11:01 | history | answered | duje | CC BY-SA 3.0 |