4b^3+4b+1=(2a-1)^2

Idril · P versus NP Offline Joined: 02 Aug 2004 Posts: 44

Prove that
$4b^3+4b+1=(2a-1)^2$ , $a,b \in \mathbb{N} \cup \left\{0\right\}$
has only solutions b=0, a=0; b=0, a=1; b=1, a=2; b=3, a=6.

**Posted:** Mon Dec 20, 2004 1:04 pm

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

nice problem from friedrich. i'm already very interestested whether he will get to the training camps this year again Very Happy

. you have any ideas how he did, idril? btw, who are you? do i know you?

so, to the problem:

we have
$(b^2+1)b=(a-1)a$ .
if $b=0$ , then $a=0$ or $a=1$ .
if $a=0$ or $a=1$ , then $b=0$ .
else we immediately get $b|a$ or $b|a-1$ .
first case: $a=kb$ . we have $k\geq 1$ . then $b^2+1=k^2b-k$ . take the discriminant to get that $k^4-4k-4$ is a square. now $(k^2-1)^2<k^4-4k-4<(k^2)^2$ for $k\geq 3$ . thus $k=1$ , giving no solution, or $k=2$ , giving $b=1$ , $a=2$ or $b=3$ , $a=6$ .
second case: $a=kb+1$ . then $b^2+1=k^2b+k$ . the discriminant now is $k^4+4k-4$ . but $(k^2)^2<k^4+4k-4<(k^2+1)^2$ for $k\geq 2$ , thus $k=1$ giving $b=1$ , $a=2$ .

Peter

**Posted:** Tue Dec 21, 2004 11:28 am

Pascual2005

i dont understand why $b|a$ or $b|a-1$ ? can you explain it please.

**Posted:** Tue Dec 21, 2004 11:50 am

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Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

... it's wrong, and i already said that before, but that post was moved since the discussion here got quite away from the problem. actually, this problem may be solved by some brutal methods for elliptic curves; this doesn't get quite nice, though.

**Posted:** Sun Dec 26, 2004 3:52 am

Pascual2005

yes, my post was here before the discussion an i have already see your reply. Thanks

**Posted:** Sun Dec 26, 2004 7:22 am

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COLOMBIA IS PASSION!!!