AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a monic polynomial with complex coefficients such that there exist a polynomial 
with complex coefficients for which ![\[P(x^2-2)=P(x)Q(x).\]](http://latex.artofproblemsolving.com/a/6/2/a62b06890b6bbcc5c9634f6de8c7ae5fe9f37828.png)
Determine all complex numbers that could be the root of .
, there exist integers 
such that, if we let 
, then the following conditions hold:
we have 
; the number 
is divisible by 
; and the numbers 
each have a prime factor not dividing .
+1 w
is irreducible for every natural number 
.
be positive real numbers satisfying![\[
a^2 + c^2 = b(a + c).
\]](http://latex.artofproblemsolving.com/9/c/0/9c0730d15b1177ee23d3f65bc2fe8b2cc85d0c62.png)
Let![\[
m = \min \left( \frac{a^2 + ab + b^2}{ab + bc + ca} \right).
\]](http://latex.artofproblemsolving.com/e/b/6/eb6f340defa4b38e35292062b202d138b9de843b.png)
Find the value of 
.
be a function such that
for all 
. Determine all such functions 
.
be real numbers such that 
Prove that 
Where 
. Find all 
such that for all 
, 
, 
and 
be points on the same circumference with 
. Let 
and 
be the projections of onto 
and 
, respectively. Prove that 
cuts the segment 
into equal parts.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
such that for any 
:
Then there are two cases either 
or 
.When we are admitting the case we cannot divide by 
. So, in the case we apply divison by 
then is a solid prerequisite to do so.Thus, 
from 
we must take the assumption in hand that . For example take the extension of the same problem in 
then the same problem reads 
,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in 
. Thankfully, we are lucky enough that we are in the field of 
.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of .
Then there are two cases either or .When we are admitting the case we cannot divide by . So, in the case we apply divison by then is a solid prerequisite to do so.Thus, from we must take the assumption in hand that . For example take the extension of the same problem in then the same problem reads ,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in . Thankfully, we are lucky enough that we are in the field of .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of ..
Then there are two cases either or .When we are admitting the case we cannot divide by . So, in the case we apply divison by then is a solid prerequisite to do so.Thus, from we must take the assumption in hand that . For example take the extension of the same problem in then the same problem reads ,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in . Thankfully, we are lucky enough that we are in the field of .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of ...The assumption whenever we get 
else we get the case .I can explain you more but the fact is I cannot use argument of field theory to explain it in total details.The reason why it's actually the case lies in field theory logics.
Which means that the roots of this equation have to be real so we can use an method that is lost in the darkness called factoring. We can factor out the from each of the terms on the left hand side and get which with more logic we can find that the possible outcomes is that if the Parantheses are 0 or the x=0 so first we can subsitute a value of x into that to make the value 0 so we get that x=5 and we finally get the solutions of and and to wrap up our proof we can prove that factoring is the best way to go because with quadratics you would only find 2 possibiliteis or 1 depending on the plus minus. And the other thing is that if you divide by x and just solve it with algebra then you will only get the solution of 5. Thus, proving that factoring is the best method out of all of them. We can use the remainder theorem to prove this which can be done easily. 
is lost? right. Well forget about field theory and all that, Let me explain it in layman's term what is actually happening. In second solution , the solution is not actually lost. The reason we are getting but not is beacuse when we are dividing by we making an assumption that and since we are making this assumption the solution is lost. For example when we divide by 
the solution is lost why 
and we are assuming 
which is equivalent to 
. Now divison by zero is not possible which is not at all very easy to explain. Now and is not possible at the same time. Thus either or . and 
can be same when we are in a field of 
. If you are avoid knowing what is field that's perfectly fine to learn later, but just in layman's term note that 
is possible in finite fields of charecteristic .
