Exponents in 2015

Algebra Level 5

Let a 1 , a_1, a 2 , a_2, a 3 , a_3, a 4 a_4 be complex numbers such that

a 1 + a 2 + a 3 + a 4 = 2 a 1 2 + a 2 2 + a 3 2 + a 4 2 = 0 a 1 3 + a 2 3 + a 3 3 + a 4 3 = 1 a 1 4 + a 2 4 + a 3 4 + a 4 4 = 5. \begin{aligned} a_1 + a_2 + a_3 + a_4 &= 2 \\ a_1^2 + a_2^2 + a_3^2 + a_4^2 &= 0 \\ a_1^3 + a_2^3 + a_3^3 + a_4^3 &= 1 \\ a_1^4 + a_2^4 + a_3^4 + a_4^4 &= 5. \end{aligned}

If a 1 6 + a 2 6 + a 3 6 + a 4 6 = m n a_1^6 + a_2^6 + a_3^6 + a_4^6 = \frac{m}{n} for some relatively prime positive integers m m and n , n, find m + n . m + n.


The answer is 25.

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Steven Yuan by Steven Yuan , 20, USA

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1 solution

Steven Yuan
Jan 24, 2015

Let s n = a 1 n + a 2 n + a 3 n + a 4 n . s_n = a_1^n + a_2^n + a_3^n + a_4^n. The conditions given in the problem become s 1 = 2 , s 2 = 0 , s 3 = 1 , s_1 = 2, s_2 = 0, s_3 = 1, and s 4 = 5. s_4 = 5.

Let the quartic polynomial x 4 + A x 3 + B x 2 + C x + D x^4 + Ax^3 + Bx^2 + Cx + D have roots a 1 , a 2 , a 3 , a 4 . a_1, a_2, a_3, a_4. From Newton's sums, we have

s 1 + A = 0 2 + A = 0 A = − 2 s 2 + A s 1 + 2 B = 0 0 − 2 ( 2 ) + 2 B = 0 B = 2 s 3 + A s 2 + B s 1 + 3 C = 0 1 − 2 ( 0 ) + 2 ( 2 ) + 3 C = 0 C = − 5 3 s 4 + A s 3 + B s 2 + C s 1 + 4 D = 0 5 − 2 ( 1 ) + 2 ( 0 ) − 5 3 ( 2 ) + 4 D = 0 D = 1 12 . \begin{aligned} s_1 + A &= 0 \\ 2 + A &= 0 \\ A &= -2 \\ & \\ & \\ s_2 + As_1 + 2B &= 0 \\ 0 - 2(2) + 2B &= 0 \\ B &= 2 \\ & \\ & \\ s_3 + As_2 + Bs_1 + 3C &= 0 \\ 1 - 2(0) + 2(2) + 3C &= 0 \\ C &= -\frac{5}{3} \\ & \\ & \\ s_4 + As_3 + Bs_2 + Cs_1 + 4D &= 0 \\ 5 - 2(1) + 2(0) - \frac{5}{3}(2) + 4D &= 0 \\ D &= \frac{1}{12}. \end{aligned}

Thus, the polynomial x 4 − 2 x 3 + 2 x 2 − 5 3 x + 1 12 x^4 - 2x^3 + 2x^2 - \frac{5}{3}x + \frac{1}{12} has roots a 1 , a 2 , a 3 , a 4 . a_1, a_2, a_3, a_4.

We want to find s 6 . s_6. First, we find s 5 . s_5. Newton's sums gives us

s 5 − 2 s 4 + 2 s 3 − 5 3 s 2 + 1 12 s 1 = 0 s 5 − 2 ( 5 ) + 2 ( 1 ) − 5 3 ( 0 ) + 1 12 ( 2 ) = 0 s 5 = 47 6 . \begin{aligned} s_5 - 2s_4 + 2s_3 - \frac{5}{3}s_2 + \frac{1}{12}s_1 &= 0 \\ s_5 - 2(5) + 2(1) - \frac{5}{3}(0) + \frac{1}{12}(2) &= 0 \\ s_5 &= \frac{47}{6}. \end{aligned}

Finally,

s 6 − 2 s 5 + 2 s 4 − 5 3 s 3 + 1 12 s 2 = 0 s 6 − 2 ( 47 6 ) + 2 ( 5 ) − 5 3 ( 1 ) + 1 12 ( 0 ) = 0 s 6 = 22 3 . \begin{aligned} s_6 - 2s_5 + 2s_4 - \frac{5}{3}s_3 + \frac{1}{12}s_2 &= 0 \\ s_6 - 2 \left (\frac{47}{6} \right ) + 2(5) - \frac{5}{3}(1) + \frac{1}{12}(0) &= 0 \\ s_6 &= \frac{22}{3}. \end{aligned}

Therefore, m + n = 22 + 3 = 25 . m + n = 22 + 3 = \boxed{25}.

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