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such that![\[(a^2 + b^2 + c^2)^2 = 3(a^3b + b^3c + c^3a) + 1. \]](http://latex.artofproblemsolving.com/9/e/4/9e400fd807a9535ede983b75dcc307d8b75a35b6.png)
with real coefficients of degree 
such that 
divides 
.
+1 w
of positive integers, such that every sum
(with 
) is a composite number, but:
for all 
; for all and 
for all 
?
denotes the greatest common divisor of 
, 
.
of positive integers, such that every sum(with ) is a composite number, but: for all ; for all and for all ? denotes the greatest common divisor of , .
, there exist sequence 
of positive integers such that 
is composite number for all 
and 
for all 
and 
for all 
, we choose sequence 
, then we choose 
where 
is prime number larger than 
and 
is a positive integer such that 
where 
is a prime divisor of 
for all 
will give us for all , 
and 
for all 
, and another bunch of odd primes (pairwise distinct and also distinct from ), say, 
. Now by virtue of CRT, choose a number 
such that:
Now set 
for 
. We claim that this works. Indeed, since the 
's are in an arithmetic progression, we have 
Now 
is always an integer, because by construction, and hence all 's are odd. For 
, the above sum is obviously composite, and for 
, the above sum reduces to 
, which is composite because 
.
, which can't be 
since we've been careful to set 
. The case 
is similar, so we win. 
and so on
then you have that sum is 
which is obviously incorrect since 
is not a composite number.
, then 
, 
and 
. Then inductively construct the 's for 
as follows.
, 
for 
and 
. This can be done by CRT as we can take sufficiently large 
's and 
's to ensure that they are pairwise coprime and also coprime to 
.
consecutive terms of a arithmetic progression whose each term is odd composite number is a composite number.
where n is odd.Clearly the sum ![$ n[x+\frac{n-1}{2}d]$](http://latex.artofproblemsolving.com/d/b/8/db8452ec6c74f166c70507213019065eb9b081c2.png)
is composite.
there is a arithmetic progression with terms such that the following holds:
.Consider the AP,
.
and common difference 
and first terms. devides GCD of 2 terms,
. devides at least one of 
or but in both cases,as 
,
.The 4th line follows from the fact that 
.
by the above 2 claims both (a),(b) holds .
for all 
, where 
.
and similarly 
.
for all , we get 
and since 
we are done!
. (Goal being )
just take 
or anything similar.
we choose primes 
such that they are all distinct and coprime to any subarray sum we have so far. Now we construct 
using 
such that:![\[ a_{n+1} \equiv -\sum_{i=j}^n a_i \mod p_{n,j}^2\]](http://latex.artofproblemsolving.com/a/1/0/a1084b5f9940fd8157deedb9a72c6b71f70534e2.png)
for all 
and![\[ a_{n+1} \equiv 1 \mod a_na_{n-1} \]](http://latex.artofproblemsolving.com/6/7/5/675b5b20c0d78964e79ebd1bf85416a31b9a40bd.png)
This is valid, since each of the squares are coprime to all elements, including 
. Since each new subarray containing is divisible by some square, they will not be prime. Additionally, is coprime to both 
and 
, so the construction is valid for 
and by induction for all , including 2016
such that:
