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When is $f(k):=8k^2+8k+1$ a square for $k\in\mathbb Z_{\geq 0}$?

How do I begin on this? I see $f(k)$ is a square for $k=0,2$, but I do not know where to go from here.

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Set your polynomial equal to $w^2,$ with $w$ a new variable. Add $1$ to both sides. What do you see? –  Will Jagy May 7 '12 at 2:47

3 Answers 3

up vote 6 down vote accepted

$y^2=8k^2+8k+1=2(2k+1)^2-1=2x^2-1$, $y^2-2x^2=-1$ is (more-or-less) a Pell's equation, concerning which there is a large and readily-available literature. One solution is $x=y=1$. If you let $(1+\sqrt2)^n=y_n+x_n\sqrt2$ for $n$ odd, then you'll have $y_n^2-2x_n^2=-1$, and in fact all solutions arise this way.

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This is a well-known integer sequence $0, 2, 14, 84, 492, 2870, 16730,\ldots$ with many interesting characterizations, see the OEIS entry A053141.

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The answer is found among Pythagorean triples $(a,b,c)$, where $c$ is perfectly square. $$8k^2+8k+1=2(2k+1)^2−1=a^2,\quad (a^2+1)/2=(2k+1)^2$$ The left side of this equation describes the "$c=b+1$" sequence of Pythagorean triples. So, your equation is square when $k =(\sqrt{c}- 1)/2$ or, when $k=(z-1)/2$.

$$\small\begin{array}{c|cccccccc} a & 7 & 41 & 239 & 1393 & 8119 & 47321 & 275807 & 1607321 \\ b & 24 & 840 & 28560 & 970224 & 32959080 & 1119638520 & 38034750624 & 1291740398520\\ c & 25 & 841 & 28561 & 970225 & 32959081 & 1119638521 & 38034750625 & 1291740398521\\ x,y & 3,4 & 20,21 & 119,120 & 696,697 & 4059,4060& 23660,23661 &37903,137904 & 803760,803761 \end{array}$$

The fascinating thing about these triples is the $(x,y,z)$ sequence from which they derive. In this sequence, $y=x+1$, and $x+y=a$, and $z^2=c=x^2+y^2$. What's more, $z=m^2+n^2$.

$$\small\begin{array}{c|cccccccc} x & 3 & 21 & 119 & 697 & 4059 & 23661 & 137903 & 803761\\ y & 4 & 20 & 120 & 696 & 4060 & 23660 & 137904 & 803760\\ z & 5 & 29 & 169 & 985 & 5741 & 33461 & 195025 & 1136689\\ m,n & 1,2 & 2,5 & 12,29 & 29,70 & 70,169 & 169,408 & 408,985 & 985,2378 \end{array}$$

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