Archive for 十月, 2008

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An Interesting Game in IMO Training Session

十月 30, 2008

I once played the following game at an IMO training session for the advanced candidates. I simply asked the students one by one, to state theorems related to number theory. This turn out to be an excellent game to get the students involved and they are really actively engaged. According to my memory, here are the theorems stated:

1) Fermat’s little theorem

2) Euler’s theorem

3) Wilson’s theorem

4) Fermat’s Last Theorem

5) Fibonacci power theorem

6) q-th power lemma

7) Fermat’s two-square theorem

8) Lagrange’s four-square theorem

9) The prime number theorem

10) Hasse-Weil bound

11) Formula for primitive pythagorean triples

12) Solution to Pell’s equation

13) Quadratic reciprocty

14) Dirichlet’s theorem on infinitude of primes in arithmetic progression

15) Iwaniec’s theorem on the largest prime of n^2 + 1

16) Chen’s theorem

17) Iwaniec’s theorem on the primes and almost primes of n^2 + 1

18) Catalan’s theorem

I think there are a few more, but I do not remember them. Please add to the list if you have interesting theorems to share.

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Diophatine Equation (1)

十月 21, 2008

The following is a classical problem, which seems to be unknown to a lot of students, including the ones in the advanced IMO training session.

Prove that y^2 = x^3 + 7 has no integer solutions.

Proof: First consider the parity of x. If x is even then we have y^2 \equiv 3 \pmod{4}, which is absurd. Hence x is odd.

If x \equiv 3 \pmod{4}, then y^2 \equiv 2 \pmod{4}, which is absurd and thus x \equiv 1 \pmod{4}.

Now write the equation as: y^2 + 1 = x^3 + 8 = (x + 2)(x^2 - 4x + 4). Since x + 2 \equiv 3 \pmod{4} There exists a prime p \equiv 3 \pmod{4} and p divides x + 2. This implies y^2 + 1 \equiv 0 \pmod{p}, and this has no solution since -1 is a square mod p iff p \equiv 1 \pmod{4} Therefore the original equation has no solution in integers.

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Answer to Polynomial Practice

十月 21, 2008

Let g(x) = (x^4 + x^3 + x^2 + x + 1) = (x - a)(x - b)(x - c)(x - d).

Then (a^2 + 1)(b^2 + 1)(c^2 + 1)(d^2 + 1) = (a + i)(a - i)(b + i)(b - i)(c + i)(c - i)(d + i)(d - i) = g(i)g(-i) = (1 - i -1 + i + 1)(1 + i -1 -i + 1) = 1(1) = 1.

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Polynomial Practice for AL Pure

十月 20, 2008

Suppose 1, a, b, c, d are roots of x^5 - 1.

Compute (1 + a^2)(1 + b^2)(1 + c^2)(1 + d^2).

Hint: There is a three-line-solution.

My solution will be given in 2 days.