10Ma5aMidtermSol

# 10Ma5aMidtermSol - Ma 1 cAn MIDTERM SOLUTIONS D...

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Unformatted text preview: Ma 1 cAn MIDTERM SOLUTIONS D. RAMAKRISHNAN (1) (a) Use induction to show that 13 divides 4 2 n +1 + 3 n +2 for every integer n ≥ . (b) List all the solutions of x 2 + y 2 ≡ m (mod 3 ), for m = 1 and m =- 1 . Solution : (a) For n = 0, 4 + 3 2 = 13, which is divisible by 13 itself. Assume by induction that the statement is true for n- 1, i.e. 4 2( n- 1)+1 + 3 ( n- 1)+2 = 4 2 n- 1 + 3 n +1 is divisible by 13. Then we have 4 2 n- 1 ≡ - 3 n +1 (mod 13). Now we have to prove that the divisibility by 13 holds for n as well. We have 4 2 n +1 + 3 n +2 ≡ 4 2 4 2 n- 1 + 3 · 3 n +1 ≡ (- 4 2 + 3)(3 n +1 ) =- 13(3 n +1 ) ≡ 0 (mod 3) . Therefore the statement holds for n when it does for n- 1. We conclude by induction that 4 2 n +1 + 3 n +2 is divisible for every integer n ≥ 0. / (b) We may take the representatives mod 3 to be 0 , 1 , 2, and identify 2 with- 1 (mod 3) when needed. Note that 0 2 = 0, 1 2 = 1, and 2 2 = 4 ≡ 1 (mod 3). Hence x 2 + y 2 can be 1 mod 3 iff exactly one of { x,y } is 0 (mod 3). So for m = 1, the solutions of x 2 + y 2 ≡ m (mod 3)are given y { ( ± 1 , 0) , (0 , ± 1) } = { (1 , 0) , (0 , 1) , (2 , 0) , (0 , 2) } . On the other hand, for x 2 + y 2 to be- 1 mod 3, we need x 2 and y 2 to be both 1 (mod 3). Thus for m =- 1, the solutions mod 3 are given by { ( ± 1 , ± 1) } = { (1 , 1) , (1 , 2) , (2 , 1) , (2 , 2) } . In either case, there are four solutions mod 3. (2) Find the group G of rigid motions of a regular tetrahedron T , and show it is a subgroup of S 4 . (By definition, T is a regular polyhedron with four equilateral triangular faces, six edges, and four vertices.) Explain your steps. Solution : Let A, B, C, D be four vertices of T . To make our calculation concrete, we may assume that coordinates of A, B, C, D in R 3 are (1 , 1 , 1), (1 ,- 1 ,- 1), (- 1 , 1 ,- 1), (- 1 ,- 1 , 1) respectively....
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10Ma5aMidtermSol - Ma 1 cAn MIDTERM SOLUTIONS D...

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