Comment: Apparently, very few people attempted the problem because most guessed that this is a Koopa Problem, which implies it’s incredibly hard…

Quite disappointed to hear that many of you failed to solve this question, especially the ones who are seeking my letter of recommendation. Anyways, here is the problem and the solution follows.

Find all solutions in integers and to:

Proof:
For or , the right hand side is , and we
easily see that the only solution is We claim that the pairs
are the only solutions. To
see this, we first note that both are primes. Now from
the fact that
for all prime , and integer , we have
for all prime , , and

Suppose Mod yields:

Now, by quadratic reciprocity, we get that the Legendre symbol

Thus is a square mod and thus
by Fermat's
little theorem. Hence the equation is now:

By quadratic reciprocity,
the equation is not solvable mod
, and hence not solvable over for the right hand side is and it's easy to see that it has no integer solution in .

Let be a point on the curve Let be the tangent line at Given intersects the curve again at Prove that

Here is good problem for my IMO students.

Given a real number . Find the greatest area of triangle ABC with circum-radius

Let Suppose , where is a rational root.

Then we have .

Multiplying both sides by yields:

Since , we must have divides

Now, from multiplying both sides by and argue similarly yields divides Hence the theorem.

Just finished my CB class tonight and I am impressed by two students. They both expressed interests in the following problem:

Suppose is a polynomial of degree . Suppose further that

(i) for all and

(ii)

Find the leading coefficient of and the degree of

I won’t discuss the solution I presented in class here. Here are two innovative ways suggested by the two students.

1) Using Taylor series. One of the students based his solution using the Taylor’s formula. The ingenious thought is that he centered the expansion at Namely, he writes the expansion as:

This method yields the correct answer and I am very impressed.

2) Using the Mean Value Theorem. Another student suggested using the Mean-Value theorem, but he could not finish the proof. I gave this idea some thought and eventually made it to work. Here is my solution whose idea is inspired by my student.

From Differentiating both sides times gives:

for all

Now applying the mean value theorem to yields:

, where

This gives:

Now note that is a polynomial with infinitely many distinct roots, namely, . [Note that they are indeed distinct because the intervals are disjoint. Therefore, by the identity theorem.

Hence, we have for all This immediately yields the degree of is , and that the leading coefficient can be found by integrating times, which yields the answer to be

For example, integrating once yields: , integrating twice yields

Likewise, integrating 101 times yields:

Book 2, pg 14. (00IQ12)(c)
(i) Let y = x^2, then we have ay^2 – by + a.
Considering the discriminant yields: b^2 – 4a^2 > 0. Therefore, it has two distinct roots: If they are both positive, then we are done. Otherwise, they must be both negative since (Note since .)
Suppose Then we have

By Viete's theorem, we have (after simplification) ), and since

(ii) This part is actually easy, just applying the previous parts. However, one must pay attention to note (and you have to explicitly state this fact) that are not equal to or Since dividing by something potentially zero is a fatal error in HKAL.

Book 4, 02IQ11.
(a) (ii), (Method 1) Here is a better way to present the solution. (to avoid adding the extra line explaining why might not be good enough.)
for all This would imply is strictly increasing for all since f has no local max or local min.

(method 2): Suppose it has more than one real root, then all roots are real since complex roots come in pairs. Let the roots be Then by Viete’s theorem, we have , and Now, This means , which is absurd.

The rest are as discussed in class.

Notes L1b, Page 14.

Given are roots of Suppose , determine the values of

Solution: We consider two cases.
Case 1, Then the roots are By Viete’s theorem, we have , which means Hence which gives

Case 2, Then we have This gives and therefore, which gives Done.

Here is my solution to the following problem given in IMO phase II training for the basic group:

Problem(1): Find all positive integer solutions to:

Proof: As I have said in l class, it is clear that . Let , where
Then we have .

Rearranging gives
Since , we have and
Therefore, Hence or

Finally, solving the three quadratics from and respectively yields is the only solution.

Problem(2) Let be a prime. Suppose is solvable. Prove that is also solvable.

Proof: First of all, we may assume , otherwise the problem is trivial. Let be a solution to Write . Then Now, since , we may write for some Therefore, if and only if i.e.

Finally, since , the linear congruence equation is solvable for all Therefore, we are done. The number is the appropriate solution.

Let

Solve the following matrix equation where

Let (E) be the following system:

Let be the augmented matrix.

Determine whether the following statements are True or False.

1) (E) can have exactly 3 solutions.

Ans: False

2) (E) has a unique solution iff

Ans: True

3) A is invertible iff

Ans: True

4) The system: is consistent if or

Ans: True

5) If , then

Ans: False

6) If , then

Ans: False

7) If (E) is consistent and , then (E) has infinitely many solutions.

Ans: True

If (E) has infinitely many solutions, then

Ans: True

9) If , then

Ans: True (be careful, this is true because A is a 3 by 3 matrix.) This is false if A is a 2 by 2 matrix.

10) If , then (E) is consistent.

Ans: False.