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Let $d,N_1,N_2\in\mathbb{Z}$ with $N_2\leq N_1(N_1-1)$, $d\in\{1,\ldots,N_2\}$ and $d\mid N_2$.

Let

$p_1(d)=N_1^2-d(N_1+3)-d^2$, and

$p_2=27N_2+2N_1^3+(-18N_1-3N_1^2)d+(9-3N_1)d^2+2d^3$.

Let $F(d)=\frac{2}{3}\sqrt{p_1(d)}\cos\left[\frac{1}{3}\arccos\left[\frac{p_2(d)}{2p_1(d)^{3/2}}\right]\right]+\frac{3+N_2+d}{3}.$ Is $F$ an increasing function w.r.t. d? The standard methods get quite messy here.

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calculus + algebra-precalculus is a contradiction... –  Aryabhata Jul 15 '11 at 18:13
    
You're right. We should probably close the question and ban this user/IP address. /. –  The Chaz 2.0 Jul 15 '11 at 18:29
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1 Answer

A hint: Forget about $N_1$, $N_2$, $d$ being integers. You only have a parameter $p \geq -{1\over4}$ replacing $N_1$ and a variable $x$ with $1\leq x\leq p^2-p$ replacing your $d$. Now plot your function $F$ as a function of $x=d$ for various values of $p$ and look at the resulting figures. If you like what you see try to prove the corresponding conjectures by analytical means.

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Hm, thanks for the response, but I really don't see how this helps. I should have said this originally, but we can also assume $d, N_1, N_2$ are positive. From the above we already have $1\leq d\leq N_2 \leq N_1(N_1-1)$ so I don't see the point in replacing $N_1$ with the parameter $p$. Letting these variables be real is obvious, but I don't see how your replacements simplify the problem. Thanks again. –  user13385 Jul 18 '11 at 12:32
    
Anything new? Thanks. –  user13385 Jul 21 '11 at 17:36
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