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+2 w
, where 
. Find
satisfying 
and![\[ a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6{,}000{,}000.\]](http://latex.artofproblemsolving.com/0/4/4/0446861fb2e63782803c4f182a529b79e213fdfd.png)
sadge
to get 
, so ![\[100(ab + bc + ca) = 2{,}000{,}000 + abc.\]](http://latex.artofproblemsolving.com/f/5/8/f5875bf64f0a266440a09bec2652d5a5c87fc823.png)
Rearrange this to 
. Ok let's try to factor this again. Add 
to both sides to get ![\[(a - 100)(b - 100)(c - 100) = 0.\]](http://latex.artofproblemsolving.com/f/d/7/fd7a08361cd4d39a4e0a4636deb54e5ba7a2d5b9.png)
. Then we just need 
, so there are 
options. Multiply this by 
, but 
is overcounted twice, so the answer is 
.
, achieved by 
and permutations for nonnegative integers 
at least 
. One can verify that these work. To show that this gives exactly 
solutions, note that if 
, then there are 
possibilities for and three possibilities to place (the possibilities won't overlap as ), so we have 
. Now if 
, then we have to add in one additional solution 
, giving in total. Now we prove must be a permutation of 
(in other words, one of the variables is ). is a solution, so now assume there exists at least two of 
that doesn't equal . WLOG 
. Note that 
.![\[a^2 b + a^2 c + b^2 a + b^2 c + c^2a + c^2 b = ab(a+b) + c^2(a+b) + c(a^2 + b^2)\]](http://latex.artofproblemsolving.com/0/4/f/04fb03668c244156ac5ad71f93c7807c9625f477.png)
, which equals![\[ ab(300 - c) + c^2 (300- c) + c((300-c)^2 - 2ab) \]](http://latex.artofproblemsolving.com/b/d/3/bd36f7ada933440517fdec53a591788feea0ded5.png)
This can be further simplified to 
. Dividing by , we get![\[ ab(100 - c) + 100 c(300 - c) = 2,000,000\]](http://latex.artofproblemsolving.com/5/4/b/54bf18cf29701c9474238ea236831f395e7e36e9.png)
This implies 
. However, we see that 
, so since 
, 
must equal 
. Now we find 
and 
(using the fact that 
). We have that 
are the roots of the quadratic![\[ X^2 - (300 - c) X + (100c - 20,000) = 0\]](http://latex.artofproblemsolving.com/7/a/0/7a0ee8888e1ae800028ce583472996eac4626461.png)
By inspection, 
is a root of this equation, and the other root is 
. Therefore one of the variables is . This means must be a permutation of 
for some .
, so ![\[ 100(ab+bc+ac)-abc = 2 \cdot 10^6. \]](http://latex.artofproblemsolving.com/5/b/b/5bb55b52bccc4e535b372771bdf84902c622269d.png)
Let 
be the monic polynomial with roots , , and 
. Then it is easy to verify given the prior equation and 
that 
. WLOG 
. Plugging into the original equation, we have that and only need to satisfy 
. If neither and are , then we obtain triples. Adding back the triple, there are 
triples.
and 
.
.
. It follows that either or 
. The second condition cannot be cyclically true for all 
(simply plot 
on a number line: the absolute difference between the middle two cannot equal the absolute difference between the outer two unless all are equal.) It follows that WLOG . Thus from the condition, we get 
for all integers 
cyclically are the only valid solutions, giving .
are the roots of the equation 
which 
. Now, we have that
, then we have 
, which shows that any ordered pairs 
will satisfy. Therefore, we need at least one of to be equal to , and the rest follows.
.
Unless one of is 100, exactly one of 
is a different sign than the others, contradicting Fermat's Last Theorem upon rearranging the equation. One of being 100 is also sufficient for equality. We can then extract .
.
and 
.
.
:
is 0. We can proceed with casework.
so we split into cases:
such that 
. Since 
.
, the equation cancels out and we get 
which is true, so as long as one variable is 100, we are fine.
, we get 
, this means that one is 0 and the other is 100.
obviously gives 
and permutations, which doesn't work.
and 
so 
. This once again implies that 
and the only way to still satisfy our case 2 is to have 
from which we get 
implying once again that 
. If none of these are divisible by 10, 
, so we get a contradiction. If one of them is divisible by 10, then one of 
.
but we have some numbers in the form 
and others in the form 
for positive integers 
. However, if we have done that, then WLOG let them be 
, but then 
, a contradiction to 
, and the same if we had 2 numbers with factors of 2, we would end up with the same contradiction that none of these will satisfy.
to both sides of the second equation, we obtain 
so , , and are roots of the equation 
Since is a root of this equation, we have 
work.
for some 
. We verify the second equation: 
as desired. such that , , and 
.
: There is clearly one case.
: If , then , so there are 
total ways here.
or something similar. The answer is .
And substituite 
to make that 
Solve this as a quadratic in 
to get or . i.e. our condition is equivalent to one of the variables being . Trivial counting gives 
From the second condition we find 
Now 
However 
is a root of this, so as long as one of 
the second condition will hold. Thus simple counting gives 
so 
,
,
.
.
.
for 
.
so they can go to Jane Street and then make 
dollars.
work
Thus we must have one of 
![\[\sum_{sym} a^2b = \sum_{cyc} ab(a + b) = \sum_{cyc} ab(300 - c) = 300 \left( \sum_{cyc} ab \right) - 3abc\]](http://latex.artofproblemsolving.com/0/1/c/01c9fdd1c151521a926b725c7efd03d13051fc05.png)
![\[= 3000000 - 30000 (a + b + c) + 300 \left( \sum_{cyc} ab \right) - 3abc - 3000000 + 9000000\]](http://latex.artofproblemsolving.com/2/6/c/26c5208457abf64d413063a02c5474a10369847a.png)
if and only if,![\[3000000 - 30000 (a + b + c) + 300 \left( \sum_{cyc} ab \right) - 3abc = 0\]](http://latex.artofproblemsolving.com/2/a/d/2ad4eeb19fa3f0a944d74e2a111922e008d4030f.png)
![\[\iff 3(100 - a)(100 - b)(100 - c) = 0\]](http://latex.artofproblemsolving.com/a/3/a/a3a0d914a216c3510c259e26990b23ee30c27f3e.png)
cases. Answer is 
by alg manipulation and Jensen's Inequality. Thus, equality cases can only occur when 
triples along with 
,