AoPS Academy
satisfying![\[f(x+f(y))=x+f(f(y))\]](http://latex.artofproblemsolving.com/8/d/a/8daa6bcb3155768ee1ba4bd22beb53a85a6e3923.png)
and 
, with the additional constraint 
.
be positive odd integers. Prove that 
doesn't divide 
.
: 
such that:
for all x,y E 
denote the number of positive integer divisors of a positive integer 
(for example, 
). Given a polynomial 
with integer coefficients, we define a sequence 
of nonnegative integers by setting![\[a_n =\begin{cases}\gcd(P(n), \tau (P(n)))&\text{if }P(n) > 0\\0 &\text{if }P(n) \leq0\end{cases}\]](http://latex.artofproblemsolving.com/7/5/9/759f4e6c8511d920b858c4234eff7a53f38ac289.png)
for each positive integer . We then say the sequence has limit infinity if every integer occurs in this sequence only finitely many times (possibly not at all). for which the sequence 
, 
, . . . has limit infinity?
.
which passes through 
and has 
as its foci. Suppose 
meets 
at 
. Without loss of generality, assume is closer to 
, and not 
(if the distances were equal, then the quadrilateral would be a rhombus and equality holds). Then by Thales' theorem, 
. Thus, since the reflection of over (which is also the reflection of in the transverse axis of the hyperbola) lies on , and 
, we have 
, which rearranges to the second part.
, and subtracting this from the square of the previous inequality, we are done.
,
is parallel to 
and the diagonals 
(b) 
.
.
,
.![\[\frac{a}{c}=\frac{b}{d}=k\]](http://latex.artofproblemsolving.com/a/f/f/affef3199d8666f19d4dbd1f26825e00a5f674d6.png)
where ![$k\in(0,1]$](http://latex.artofproblemsolving.com/e/3/5/e355e6cf9e4ffd5a1a224da68aeac92e9701e769.png)
.
and 
,![\[\begin{gathered}(1-k^{2})^{2}\ge0\\ c^{2}(1-k^{2})d^{2}(1-k^{2})\ge0\\ (c^{2}-a^{2})(d^{2}-b^{2})\ge0\\ a^{2}b^{2}+c^{2}d^{2}\ge a^{2}d^{2}+b^{2}c^{2}\\ a^{2}b^{2}+c^{2}d^{2}+(a^{2}c^{2}+b^{2}d^{2})\ge a^{2}d^{2}+b^{2}c^{2}+(a^{2}c^{2}+b^{2}d^{2})\\ (a^{2}+d^{2})(b^{2}+c^{2})\ge(a^{2}+b^{2})(c^{2}+d^{2})\\ AD^{2}\cdot BC^{2}\ge AB^{2}\cdot CD^{2}\\ AD\cdot BC\ge AB\cdot CD \end{gathered}\]](http://latex.artofproblemsolving.com/8/3/0/830557f7f74efcc0fcd96fffbdf5365f243d9fca.png)
(b) From part (a) we have ![\[\begin{gathered}AD\cdot BC\ge AB\cdot CD\\ 2AD\cdot BC\ge2AB\cdot CD\\ (a^{2}+b^{2}+c^{2}+d^{2})+2AD\cdot BC\ge2AB\cdot CD+(a^{2}+b^{2}+c^{2}+d^{2})\\ (a^{2}+d^{2})+2AD\cdot BC+(b^{2}+c^{2})\ge(a^{2}+b^{2})+2AB\cdot CD+(c^{2}+d^{2})\\ AD^{2}+2AD\cdot BC+BC^{2}\ge AB^{2}+2AB\cdot CD+CD^{2}\\ (AD+BC)^{2}\ge(AB+CD)^{2}\\ AD+BC\ge AB+CD \end{gathered}\]](http://latex.artofproblemsolving.com/e/1/a/e1aef69883ab361eed4573211f69eeaefad4fbd3.png)
