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that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
to both sides to get ![\[\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^2-11x\]](http://latex.artofproblemsolving.com/8/d/3/8d3b681fb573fb513687dc58f154bc4a8bc61731.png)
Divide through x to get ![\[\frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}=x-11\]](http://latex.artofproblemsolving.com/c/0/3/c032b3a71f5fa778337b129455e765b96f20d2a9.png)
Adding the two outside and two inside terms, the left hand side is ![\[\frac{2x-22}{x^2-22x+57}+\frac{2x-22}{x^2-22x+85}\]](http://latex.artofproblemsolving.com/2/f/a/2fa8a0b4f87c9768c305de47d7223bdeb6fce8e3.png)
Therefore the equation becomes ![\[\frac{1}{x^2-22x+57}+\frac{1}{x^2-22x+85}=\frac{1}{2}\]](http://latex.artofproblemsolving.com/d/e/1/de155fb023956d3726e6d950a7f28575bb487856.png)
Let 
, so this is ![\[\frac{1}{y}+\frac{1}{y+28}=\frac{1}{2}\]](http://latex.artofproblemsolving.com/a/c/8/ac89851145abdde1c0604f6f73ea381269e80592.png)
This clears to ![\[y^2+24y-56=0\]](http://latex.artofproblemsolving.com/2/9/8/298f204fbeeeeda8019a1237b9474ee74f4d85fd.png)
which solves to be 
(but I somehow got negative that on the AIME). Then 
gives 
(Pedantic and irrelevant edit: by this, I mean the four possible combinations of 
and 
, no just the two possibilities of 
and 
), so taking the obvious maximum we get 
wrong. 
and then pairing terms on the left to get denominators of 
and 
That immediately led to factoring out a 
and then I also discovered that I could factor out 
. That left a quadratic for 
to solve by the quadratic formula.
So I must have made an arithmetic error very near the final step.
. That HAD to be important. It probably meant that 
, for example. There's no other possible value for 
that would even make sense. But I wasn't sure what was up with the right hand side. The left-hand side looked natural, but the 
seemed completely haphazardly thrown. The problem was hiding something, but I couldn't see what. was almost certain, I set 
and began crunching. Pairing, rearranging, whatever I could do to try to knock this out.
, 
, 
to cancel, but no such luck). After some minutes of this I accepted a temporary defeat and decided to come back later. I proceeded to squash the other eight problems I hadn't attempted yet within an hour, then returned to the sole survivor.
only covered four roots. So there had to be other roots left. I wondered if a quadratic factored out of the thing somehow, with some strange irrational roots.
to see what the constant term might be under a full expansion. This gives 
... which gives precisely the right-hand side. Oops! I swore audibly in the testing room. At least I knew what the 
there was for now. is a root, that means I could annihilate a factor of 
from both sides. That caused me to add the to both sides, and then kill the 's. This left ![\[ \frac{1}{x-3} + \frac{1}{x-5} + \frac{1}{x-17} + \frac{1}{x-19} = x-11. \]](http://latex.artofproblemsolving.com/9/4/4/9448b46d69d9a93fa0487d4545a460eec9a99614.png)
Now I did the same substitution as before, and at this point the fractions paired off nicely to wipe out the from both sides. Indeed, you just get ![\[ \frac{2r}{r^2-64} + \frac{2r}{r^2-36} = r \]](http://latex.artofproblemsolving.com/4/8/d/48d9b8527b825420d1d7046b1d0dbd57b764af49.png)
which immediately drops to a quadratic in 
. And that was straightforward!
and you didn't notice?
reminds one of the fact that 
(only valid for 
, but we can draw a veil over that for now ). Indeed, since 
, then letting 
yields 
, and similarly for the rest. Therefore, the equation becomes ![\[-\left(1+\frac{x}{3}+\left(\frac{x}{3}\right)^2+\cdots\right)-\left(1+\frac{x}{5}+\left(\frac{x}{5}\right)^2+\cdots\right)\]](http://latex.artofproblemsolving.com/8/6/a/86ad88afc2420c8352f9993fd7807c008d51bb6e.png)
![\[-\left(1+\frac{x}{17}+\left(\frac{x}{17}\right)^2+\cdots \right)-\left(1+\frac{x}{19}+\left(\frac{x}{19}\right)^2+\cdots\right)\]](http://latex.artofproblemsolving.com/6/e/9/6e9cab4eaaca27f6904b9ec612c13b7bff2f7e03.png)
![\[=x^2-11x-4.\]](http://latex.artofproblemsolving.com/7/9/a/79ac2055cab0f6233d5bea64d604f220df8246dc.png)
It is now clear to add to both sides (by evenly distributing it among the terms on the LHS), because they all have a constant term of 
, and 
, which is the constant term on the RHS. Therefore, adding to both sides we get ![\[-\left(\frac{x}{3}+\left(\frac{x}{3}\right)^2+\cdots\right)-\left(\frac{x}{5}+\left(\frac{x}{5}\right)^2+\cdots\right)\]](http://latex.artofproblemsolving.com/4/b/c/4bceafc58c73cdd3f99920f652e6dc1580038ab8.png)
![\[-\left(\frac{x}{17}+\left(\frac{x}{17}\right)^2+\cdots \right)-\left(\frac{x}{19}+\left(\frac{x}{19}\right)^2+\cdots\right)\]](http://latex.artofproblemsolving.com/9/8/1/98118213d1da31a71bf668fb2fc39d778ca9b516.png)
![\[=x^2-11x.\]](http://latex.artofproblemsolving.com/9/8/6/986d9a0ff599171eda50920300739cb3ff2c7356.png)
The rest is easy. (Divide through by , switch back to closed-forms, transform, etc.)
to each fraction yields
to get rid of the terms in the numerators of the RHS;
and combining the second and third terms yields 
This is helpful because it allows us to substitute later on.
(similar step done previously) to clean up the numerators, we have
we have
or 
Since the problem asks for the largest root, we can ignore 
Substituting 
into we have
and since the problem asks for the largest root, it is obvious that 
so 
to both sides? Was scrolling through thread and caught that it was first step, but how would I figure that out on the actual test?
, so you instinctively add. The trick is that you notice how it distributes to all the fractions in the first place.
on the RHS, which is compatible with our initial substitution.
, and notice that
Now let 
, such that 
Now we want the largest solution, which is 
We can extract the answer of 
.
Unfortunately, I got it wrong 'cause I can't multiply.
satisfying![\[ \frac{3}{x-3} + \frac{5}{x-5} + \frac{17}{x-17}
+ \frac{19}{x-19} = x^2-11x-4. \]](http://latex.artofproblemsolving.com/e/4/3/e435160bea65a47d274bdd6b97f3d9710125aa29.png)
to both sides, we obtain
and assuming 
(we will find a larger solution), dividing by yields
upon which diving by 
(assuming 
) and substituting 
yields
Since we want the largest real number 
we take the positive root and finish: 
Thus, the largest value of is 
Extract 
s into 
s so yeah
This is well motivated because it's very common when there are fractions with pairwise sums to be the same that we combine denominators of some to manipulate. From here, just substitute 
(you can choose any c you want, it'll work) and solving for y solves the problem.
to both sides and divide by to get
![\[\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x-17} + \frac{1}{x-19} = x - 11\]](http://latex.artofproblemsolving.com/e/0/5/e057eafb965022667b3ed654eb514d72d42b22c8.png)
and then cross multiplying and dividing by 
gives![\[\frac{2}{(x-3)(x-19)} + \frac{2}{(x-5)(x-17)} = 1\]](http://latex.artofproblemsolving.com/2/2/b/22b6df97f35fe388e7ba464edefcb7657d03e348.png)
Then let 
, so we have![\[\frac{2}{(y^2 - 64)} + \frac{2}{(y^2 - 36)} = 1 \implies \]](http://latex.artofproblemsolving.com/f/9/d/f9dfed180b2d6538104b2cfc41125cd9b3e806f8.png)
![\[4y^2 - 200 = (y^2 - 64)(y^2 - 36)\]](http://latex.artofproblemsolving.com/e/7/3/e73a6b1d44290be1c47bd9b71e88ff5e8aaf9fc8.png)
Now, let 
.![\[4z - 200 = (z - 64)(z - 36) \implies z^2 - 104z + 2504 = 0 \implies z = 52 + 10\sqrt{2}\]](http://latex.artofproblemsolving.com/5/d/2/5d28884524a6352f3b4c0b717e4b6afdf2fc2b13.png)
is our answer.
be the largest real solution to the equation ![\[\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}= x^2-11x-4.\]](http://latex.artofproblemsolving.com/e/8/7/e87fdc9fe63d29a46e57dfb92c2d0448441b4b55.png)
There are positive integers 
such that 
.
.
.
![\[1+\frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}= x^2-11x\]](http://latex.artofproblemsolving.com/4/8/3/4838269f54d0a97309224541f95667ed721e766f.png)
![\[\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}= x^2-11x\]](http://latex.artofproblemsolving.com/7/3/7/737641d53dfd476afe13c3bd59357398561bae33.png)
![\[\frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}= x-11\]](http://latex.artofproblemsolving.com/4/a/8/4a8f6eb856c7bf47bc2abeb151cc6a7a79b9be96.png)
,
:![\[\frac{1}{y-8} + \frac{1}{y-6} + \frac{1}{y+6} + \frac{1}{y+8} = y\]](http://latex.artofproblemsolving.com/6/3/0/630fe2a7b65524c9c0c29b5f39b06b6eb5dad08f.png)
![\[\frac{(y+8)+(y-8)}{(y+8)(y-8)} + \frac{(y+6)+(y-6)}{(y+6)(y-6)} = y\]](http://latex.artofproblemsolving.com/8/7/b/87b1367c5b7cae100067974ab2f1d711c084be17.png)
![\[\frac{2y}{y^2-64} + \frac{2y}{y^2-36} = y\]](http://latex.artofproblemsolving.com/1/8/0/180c55e086edfaba499db2233ca425e04841f66e.png)
![\[\frac{2}{y^2-64} + \frac{2}{y^2-36} = 1\]](http://latex.artofproblemsolving.com/9/1/b/91b623bff277fb156bfe95398913ae40f519caae.png)
and expand:![\[2y^2-128+2y^2-72=y^4-100y^2+2304\]](http://latex.artofproblemsolving.com/a/e/5/ae52631c7df8cd0e37e760d7a39077ce0b4b87cd.png)
![\[y^4-104y^2+2504=0\]](http://latex.artofproblemsolving.com/c/c/6/cc64e85796265bb331309f45ea7c194fb994abc3.png)
and 
.
,
,
.
,
.
![\[\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4\]](http://latex.artofproblemsolving.com/f/6/d/f6da4c5e2af3e6b71da3837094a15075ff02ba72.png)
![\[4+\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x\]](http://latex.artofproblemsolving.com/7/8/3/783418b099fa8af883e285d7b9c822966db1cbf8.png)
![\[(1+\frac{3}{x-3})+(1+\frac{5}{x-5})+(1+\frac{17}{x-17})+(1+\frac{19}{x-19})=x^2-11x\]](http://latex.artofproblemsolving.com/6/1/b/61ba6fc5281f16ce363012f935b6ce2f82321fa1.png)
.![\[\frac{1}{a+8}+\frac{1}{a+6}+\frac{1}{a-6}+\frac{1}{a-8}=a\]](http://latex.artofproblemsolving.com/3/9/e/39e03e7ccbec17c0c5c33545eb579ec188299685.png)
![\[\frac{2a}{a^2-64}+\frac{2a}{a^2-36}=a\]](http://latex.artofproblemsolving.com/4/d/e/4de97279adedc938d38a97e192b9f7e24ee7badf.png)
![\[\frac{1}{a^2-64}+\frac{1}{a^2-36}=\frac{1}{2}\]](http://latex.artofproblemsolving.com/8/3/b/83b4467df8f742a24474a5aa660d87002c711a82.png)
![\[\frac{2a^2-100}{a^4-100a^2+2304}=\frac{1}{2}\]](http://latex.artofproblemsolving.com/0/2/a/02accf1bd0ebed55adc5dc10e47d96e83c02d49a.png)
![\[a^4-104a^2+2504=0\]](http://latex.artofproblemsolving.com/3/b/6/3b6e430f20b1a8b196d1ee93850a55e70ed3af14.png)
![\[\implies a^2 = \frac{104\pm\sqrt{104^2-4(2504)}}{2}\implies \frac{104\pm\sqrt{800}}{2}\]](http://latex.artofproblemsolving.com/e/2/c/e2cdf3b2af60683af12931f80675744af7daf08d.png)
![\[\implies \frac{104\pm\ 20 \sqrt{10}}{2}\implies 52+10\sqrt{2} \implies a=\sqrt{52+\sqrt{200}}\]](http://latex.artofproblemsolving.com/9/b/4/9b40cbacbe66eb1d9c1ef7133129599dbcd047d4.png)
Which gives us that 
![\[\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4 \]](http://latex.artofproblemsolving.com/d/2/3/d23c8b4cf49dd25de9803d629b665ed23707f406.png)
on the top and a 
on the bottom of the fraction 
,![\[\dfrac{3}{x-3}+1-1=\dfrac{x-3+3}{x-3}-1=\dfrac{x}{x-3}-1\]](http://latex.artofproblemsolving.com/7/e/a/7ea62a84cec98b96f059cc46fd9a38efed103a73.png)
.![\[\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}-4=x^2-11x -4\]](http://latex.artofproblemsolving.com/4/2/a/42aaf74d3613476e2aea10974ef52238667ea2ab.png)
The ![\[\dfrac{1}{x-3}+\dfrac{1}{x-19}+\dfrac{1}{x-5}+\dfrac{1}{x-17}=\dfrac{2x-22}{(x-3)(x-19)}+\dfrac{2x-22}{(x-5)(x-17)}=x-11\]](http://latex.artofproblemsolving.com/a/1/6/a160f5f9ffddda2125b4517b4fdcdecfef965c34.png)
'![\[\dfrac{1}{(x-3)(x-19)}+\dfrac{1}{(x-5)(x-17)}=\dfrac{1}{2}\]](http://latex.artofproblemsolving.com/6/1/0/61088096b2ccc6e0edee4a115fb52453f0944d81.png)
![\[2((x-5)(x-17)+(x-3)(x-19))=(x-3)(x-5)(x-17)(x-19)\]](http://latex.artofproblemsolving.com/1/5/3/153721dd7534f33f009edfc29397636aa18fcdf4.png)
and a 
term, which made me think of substitution.
,
.
.
.
.
,
,
for all 
![\[\frac{3}{P_3}+\frac{5}{P_5}+\frac{17}{P_{17}}+\frac{19}{P_{19}}=x^2-11x-4.\]](http://latex.artofproblemsolving.com/d/a/b/dabc2c10c06900378bdf900d735581a91ecb8b56.png)
As in earlier solutions, we pull over the ![\[\frac{1}{P_3}+\frac{1}{P_5}+\frac{1}{P_{17}}+\frac{1}{P_{19}}=x-11.\]](http://latex.artofproblemsolving.com/d/9/0/d902f1134a711d3a01b31f6dab9a9be7fc907445.png)
Here's where we do something mindless (but it works): time to clear the denominators.![\[P_3P_5P_{17}+P_3P_5P_{19}+P_3P_{17}P_{19}+P_5P_{17}P_{19}=(x-11)P_3P_5P_{17}P_{19}.\]](http://latex.artofproblemsolving.com/7/c/4/7c4d1cc2f5ad841de3884b5292173010f883b382.png)
Now we want to get rid of that ![\[P_5P_{17}(2x-22)+P_3P_{19}(2x-22)=(x-11)P_3P_5P_{17}P_{19}\]](http://latex.artofproblemsolving.com/d/1/f/d1f2f112b223fe4424067636fbd8bc34a4097f8a.png)
, so ![\[2P_5P_{17}+2P_3P_{19}=P_3P_5P_{17}P_{19}\]](http://latex.artofproblemsolving.com/0/0/3/003f3b9bff9fd1355e1b9a3e3429a06a9daec3eb.png)
![\[(P_3P_{19}-2)(P_5P_{17}-2)=4.\]](http://latex.artofproblemsolving.com/0/7/4/074c7426b52bb4b421994d93f052ce1e64d18d8e.png)
Now we rid ourselves of the shorthand and expand inside the parentheses to get![\[(x^2-22x+55)(x^2-22x+83)=4.\]](http://latex.artofproblemsolving.com/4/9/f/49ff5f92234b2bd9481cf21a545aad52eca543a2.png)
Instead of substituting, we use difference of squares to get![\[(x^2-22x+69)^2=4+14^2=200\]](http://latex.artofproblemsolving.com/7/c/6/7c61da251c8e735787a23db92770efb0d54d3c05.png)
, hence ![\[x^2-22x+69=\pm\sqrt{200}\]](http://latex.artofproblemsolving.com/0/3/9/039a564284ab5a64a274fc142f75aa1c38470935.png)
. Completing the square,![\[(x-11)^2=52\pm\sqrt{200}\]](http://latex.artofproblemsolving.com/8/9/e/89ea5246c6792609661b458d783e52a12c04d6c0.png)
, so ![\[x=11\pm\sqrt{52\pm\sqrt{200}}\]](http://latex.artofproblemsolving.com/e/3/0/e3059b6d5362ef4155db03eb194f96745da16a4e.png)
(four roots),
.
Why couldn't have the form of the answer be different...
to both sides gives 
.
.
,
.
.
.
,
.
.
and 
.
.
Since 
Grouping the fractions, we get
So either 
or:
If we let 
Substituting, we have
Therefore, 
,
to get 
is the answer.
.
Let 
Let 
.
Then 
and 
.
,
.
In other words, if we add 
Dividing both sides by 
Now, we obtain 
Dividing both sides by 
It is not hard to solve this for 
to obtain 
Out of the 
is the largest. This is also larger than 
so this is our answer.
![\[\frac{1}{x-3}+\frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}=x-11.\]](http://latex.artofproblemsolving.com/d/d/9/dd929a63ec5eeb769776ccc133a9999aaeb2273c.png)
This is equivalently![\[\frac{2x-22}{x^2-22x+57}+\frac{2x-22}{x^2-22x+85}=x-11.\]](http://latex.artofproblemsolving.com/5/1/3/513a46ede323d2074f0126f42356818e8b6aed62.png)
and substitute it into the expression to make things a bit nicer.![\[\frac{2t}{t^2-64}+\frac{2t}{t^2-36}=t.\Longleftrightarrow \frac{2}{t^2-64}+\frac{2}{t^2-36}=1\]](http://latex.artofproblemsolving.com/a/2/6/a26e70ba944ee3d1576251c74f1178a7c316b814.png)
(sidenote: don't forget that 
.
Hence, 
is the largest solution 
,
degree polynomial, we substitute 
,
.
,![\[\frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^2-11x.\]](http://latex.artofproblemsolving.com/1/1/b/11bad9f132369b1d278549095edaf534ab2a20e3.png)
Dividing both sides by 
,
.![\[\frac{1}{r+8}+\frac{1}{r+6}+\frac{1}{r-6}+\frac{1}{r-8}=r,\]](http://latex.artofproblemsolving.com/7/5/e/75ebae7c5e92bfb0adf941badb874ee7a5647078.png)
so![\[\frac{2r}{r^2-64}+\frac{2r}{r^2-36}=r.\]](http://latex.artofproblemsolving.com/8/b/9/8b97d02c6997ce3e8ce0c8ca73214fe374a96963.png)
Hence,![\[\frac{2}{r^2-64}+\frac{2}{r^2-36}=1,\]](http://latex.artofproblemsolving.com/a/a/c/aacc213bf95970f13c36455a0eee1e615432ecb4.png)
so we can simplify this to a quadratic in 
,
.
.
,
is a root. We let 
.
.
to get 
.