Difference between revisions of "KGS math club/solution 2 1"

Revision as of 22:03, 30 September 2014

Answer: No.

Proof:

Lemma 1: 2007 = 223 * 3^2  (trivial)
Lemma 2: x(n) = 2^n * (n^2 + 1)  (letting x(0) = 1, x(1) = 4 and x(2) = 20)
	Proof: Induction over n.
	For n = 3: n(3) = 6 * 20 - 12 * 4 + 8 = 80 = 2^3 * (3^2 + 1)
	For n > 3: assume x(n) = 2^n * (n^2 + 1) and similarly for x(n-1) and x(n-2).
		then x(n+1) = 6 * 2^n * (n^2 + 1) - 12 / 2 * 2^n * (n^2 - 2n + 2) + 
                               8 / 4 * 2^n * (n^2 - 4n + 5)
			= 2^n * ( (6 - 6 + 2) n^2 + (-6*-2 + 2*-4) n + (6 - 6*2 + 2*5) )
			= 2^n * 2 * (n^2 + 2n + 1 + 1)
			= 2^(n+1) * ( (n+1)^2 + 1)
	Thus, formula holds for all n >= 3, q.e.d.

Serious business observation: neither 3 nor 223 (both odd) are going to divide 2^n
(which is always even) given any n, same applies the other way around.

Thus it remains to show that neither 3 not 223 divide n^2+1, the other factor of x(n).

That is equailent to showing that
For all n, n^2 != 2 (mod 3) and n^2 != 222 (mod 223)
or in other words, that 2 and 222 are not quadratic residues modulo 3 and 223 respectively.

Using Euler's criterion (for those _really_ interested in the proof, look it up in Wikipedia) we decide:
2^((3-1)/2) = 2^1 = 2 = -1 (mod 3) meaning that 2 is _not_ a quadratic residue modulo 3
and
222^((223-1)/2) = 222^111 = 222 = -1 (mod 223) meaning that, indeed, 222 is not a quadratic residue modulo 223

concluding the proof. Thank YOU for reading this far. Looks like that makes two of us (not having life...)

Yours, sigs

Comments

The idea behind inductive formula:

The recursion $x_n=6x_{n-1}-12x_{n-2}+8x_{n-3}$ can be rewritten more symmetrically as $x_n-4x_{n-1}+4x_{n-2}=2(x_{n-1}-4x_{n-2}+4x_{n-3})$ which when applied (n-2) times gives $x_n-4x_{n-1}+4x_{n-2}=2^{n-2}(x_2-4x_1+4x_0)=2^{n-2}(20-4*4+4*1)=2^{n+1}.$

This new recursion can be rewritten as $x_n-2x_{n-1}=2^{n+1}+2(x_{n-1}-2x_{n-2})$

and (n-2) repeated applications gives $x_n-2x_{n-1}=(n-1)2^{n+1}+2^{n-1}(x_1-2x_0)=(n-1)2^{n+1}+2^{n}=(2n-1)2^{n}$

Finally, repeated applications of this new recursion $x_n=(2n-1)2^{n}+2x_{n-1}$ gives $x_n=2^{n}((2n-1)+(2n-3)+(2n-5)+\cdots+3+1)+2^nx_0=2^{n}(n^2+1)$

The idea behind the quadratic residues step:

$x^2+1=0 Mod(p)$ has a solution iff $p=1Mod(4)$ (where $p$ is Prime). A result of Euler states that $x^{\frac{p-1}{2}}=\pm 1$ depending on whether x is a quadratic residue. If $p=1Mod(4)$ , then $x^{\frac{p-1}{2}}=+1$ .

@@ Line 51: / Line 51: @@
 The idea behind the quadratic residues step:
-<math>x^2+1=0 Mod(p)</math> has a solution iff <math>p=1Mod(4)</math> (<math>p</math> is Prime). To check the if part, note that squares are either <math>0 Mod(4) </math> or <math>1 Mod(4)</math> , and hence <math>x^2+1</math> is either <math>1Mod(4)</math> or <math>2Mod(4)</math>. Since <math>3</math> and <math>223</math> are <math>3Mod(4)</math>, there is no solution to <math>x^2+1=0Mod(3 or 223)</math>.
+<math>x^2+1=0 Mod(p)</math> has a solution iff <math>p=1Mod(4)</math> (where <math>p</math> is Prime). A result of Euler states that <math>x^{\frac{p-1}{2}}=\pm 1</math>
+depending on whether x is a quadratic residue. If <math>p=1Mod(4)</math>, then <math>x^{\frac{p-1}{2}}=+1</math>.

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Difference between revisions of "KGS math club/solution 2 1"

Revision as of 22:03, 30 September 2014

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