But it's so large!

Algebra Level 5

{ a + b + c + d = 3 a 2 + b 2 + c 2 + d 2 = 5 a 3 + b 3 + c 3 + d 3 = 3 a 4 + b 4 + c 4 + d 4 = 9 \large{ \begin{cases} {a+b+c+d=3} \\ {a^2+b^2+c^2+d^2=5} \\ {a^3+b^3+c^3+d^3=3} \\ {a^4+b^4+c^4+d^4=9} \end{cases} }

Given that a , b , c a,b,c and d d are complex numbers that satisfy the equation above. If a 2015 + b 2015 + c 2015 + d 2015 a^{2015} + b^{2015} + c^{2015} + d^{2015} can be written as p q + p r 1 p^q + p^r - 1 for positive integers p , q , r p,q,r , evaluate p + q + r + 1 p+q+r+1 .


The answer is 3026.

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2 solutions

Daniel Liu
Jun 28, 2015

Let a , b , c , d a,b,c, d be the roots of the quartic x 4 + k 3 x 3 + k 2 x 2 + k 1 x + k 0 x^4+k_3x^3+k_2x^2+k_1x+k_0 .

Define a n + b n + c n + d n = P n a^n+b^n+c^n+d^n=P_n .

By Newton's sums, we see that: { P 1 + k 3 = 0 P 2 + k 3 P 1 + 2 k 2 = 0 P 3 + k 3 P 2 + k 2 P 1 + 3 k 1 = 0 P 4 + k 3 P 3 + k 2 P 2 + k 1 P 1 + 4 k 0 = 0 \left\{ \begin{array}{l} P_1+k_3=0\\ P_2+k_3P_1+2k_2=0\\ P_3+k_3P_2+k_2P_1+3k_1=0\\ P_4+k_3P_3+k_2P_2+k_1P_1+4k_0=0 \end{array}\right. Substituting our known values in: { 3 + k 3 = 0 5 + 3 k 3 + 2 k 2 = 0 3 + 5 k 3 + 3 k 2 + 3 k 1 = 0 9 + 3 k 3 + 5 k 2 + 3 k 1 + 4 k 0 = 0 \left\{ \begin{array}{l} 3+k_3=0\\ 5+3k_3+2k_2=0\\ 3+5k_3+3k_2+3k_1=0\\ 9+3k_3+5k_2+3k_1+4k_0=0 \end{array}\right. Solving the system, we get ( k 3 , k 2 , k 1 , k 0 ) = ( 3 , 2 , 2 , 4 ) (k_3, k_2, k_1, k_0)=(-3, 2, 2, -4) Thus, the polynomial with roots a , b , c , d a, b, c, d is x 4 3 x 3 + 2 x 2 + 2 x 4 x^4-3x^3+2x^2+2x-4

Applying the Rational Root Theorem, we find that x = 1 , 2 x=-1, 2 are roots of the quartic, and long division gives us that x 4 3 x 3 + 2 x 2 + 2 x 4 = ( x + 1 ) ( x 2 ) ( x 2 2 x + 2 ) x^4-3x^3+2x^2+2x-4=(x+1)(x-2)(x^2-2x+2) Then we apply the quadratic equation and find that the remaining two roots are 1 ± i 1\pm i .

So the roots are ( a , b , c , d ) = ( 1 , 2 , 1 + i , 1 i ) (a,b,c,d)=(-1, 2, 1+i, 1-i) . Now we can directly compute P 2015 P_{2015} :

P 2015 = ( 1 ) 2015 + 2 2015 + ( 1 + i ) 2015 + ( 1 i ) 2015 = 1 + 2 2015 + 2 2015 ( 1 2 + 1 2 i ) 2015 + 2 2015 ( 1 2 1 2 i ) 2015 = 1 + 2 2015 + 2 2015 cis ( π 4 ) 2015 + 2 2015 cis ( π 4 ) 2015 = 1 + 2 2015 + 2 2015 cis ( 2015 π 4 ) + 2 2015 cis ( 2015 π 4 ) = 1 + 2 2015 + 2 2015 cis ( π 4 ) + 2 2015 cis ( π 4 ) = 1 + 2 2015 + 2 1007 2 ( 1 2 1 2 i ) + 2 1007 2 ( 1 2 + 1 2 i ) = 1 + 2 2015 + 2 1007 ( 1 i ) + 2 1007 ( 1 + i ) = 2 2015 + 2 1008 1 \begin{aligned} P_{2015}&=(-1)^{2015}+2^{2015}+(1+i)^{2015}+(1-i)^{2015}\\ &= -1+2^{2015}+\sqrt{2}^{2015}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i\right)^{2015}+\sqrt{2}^{2015}\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}i\right)^{2015}\\ &= -1+2^{2015}+\sqrt{2}^{2015}\text{cis}\left(\dfrac{\pi}{4}\right)^{2015}+\sqrt{2}^{2015}\text{cis}\left(-\dfrac{\pi}{4}\right)^{2015}\\ &= -1+2^{2015}+\sqrt{2}^{2015}\text{cis}\left(\dfrac{2015\pi}{4}\right)+\sqrt{2}^{2015}\text{cis}\left(-\dfrac{2015\pi}{4}\right)\\ &= -1+2^{2015}+\sqrt{2}^{2015}\text{cis}\left(-\dfrac{\pi}{4}\right)+\sqrt{2}^{2015}\text{cis}\left(\dfrac{\pi}{4}\right)\\ &= -1+2^{2015}+2^{1007}\cdot \sqrt{2}\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}i\right)+2^{1007}\cdot \sqrt{2}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i\right)\\ &= -1+2^{2015}+2^{1007}\cdot(1-i)+2^{1007}\cdot(1+i)\\ &= 2^{2015}+2^{1008}-1 \end{aligned} So ( p , q , r , s ) = ( 2 , 2015 , 1008 , 1 ) (p, q, r, s)=(2, 2015, 1008, 1) and our final answer is 2 + 2015 + 1008 + 1 = 3026 2+2015+1008+1=\boxed{3026}

Ohhh, I didn't refresh my page and I didn't see that you already post a solution. By the way, you need to show that p p is unique.

Pi Han Goh - 5 years, 11 months ago

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Well, ( 1008 , 2015 ) = 1 (1008, 2015)=1 so it is unique.

Daniel Liu - 5 years, 11 months ago

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Note that if the value of s s is not stated, then we could have multiple solutions that depend on s s .

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin What is the other solution for s s another s = 1 s=1 sir?

uzumaki nagato tenshou uzumaki - 5 years, 11 months ago

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@Uzumaki Nagato Tenshou Uzumaki For example, we could have p = 3 , q = 3000 , r = 3 p = 3, q = 3000, r = 3 and s = 3 3000 + 3 3 2 2015 2 1008 s = 3^{3000} + 3^3 - 2^{2015} - 2^{1008} . Now, clearly numerous other combinations would work, and the sum will be ridiculous.

This is why I edited your statement to indicate that s = 1 s = 1 . There is still the work of showing that we have a unique answer. As pointed out, this is equivalent to gcd ( 1008 , 2015 ) = 1 \gcd(1008, 2015 ) = 1 .

Calvin Lin Staff - 5 years, 11 months ago

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@Calvin Lin Thanks you sir i haven't think so far. And thanks too for corect that :)

uzumaki nagato tenshou uzumaki - 5 years, 11 months ago

same with my solution . good but u have to show that ( 1 2 + 1 2 i ) 2015 = 2 1007 2 2015 ( 1 2 1 2 i ) \left ( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i \right )^{2015} = \frac{2^{1007}}{\sqrt{2}^{2015}} \left ( \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i \right ) from De' Moivre or we can use ( 1 + i ) 4 = 4 (1+i)^{4} = -4

uzumaki nagato tenshou uzumaki - 5 years, 11 months ago

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Hello, I have edited my solution. Is it clear now?

Daniel Liu - 5 years, 11 months ago

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perfecto thanks you ::)

uzumaki nagato tenshou uzumaki - 5 years, 11 months ago
Pi Han Goh
Jun 28, 2015

Consider the expansion ( a + b + c + d ) 2 = ( a 2 + b 2 + c 2 + d 2 ) + 2 ( a b + a c + a d + b c + b d + c d ) (a+b+c+d)^2 = (a^2+b^2+c^2+d^2) + 2(ab+ac+ad+bc+bd+cd) , upon substitution, we get a b + a c + a d + b c + b d + c d = 3 2 5 2 = 2 ab+ac+ad+bc+bd+cd = \frac{3^2 - 5}2 = 2 .

Since the values of a , b , c a,b,c and d d satisfy these 4 equations and are interexchangeable, then there exist a 4th degree polynomial that have roots a , b , c a,b,c and d d . By Vieta's formula, the polynomial must be

x 4 3 x 3 + 2 x 2 + m x + n x^4 - 3x^3 +2x^2 + mx + n

for unique constants m m and n n . Take its sum:

( x 4 3 x 3 + 2 x 2 + m x + n ) = 0 x 4 3 x 3 + 2 x 2 + m x + n 1 = 0 9 3 ( 3 ) + 2 ( 5 ) + m ( 3 ) + n ( 4 ) = 0 3 m + 4 n = 10 ( ) \begin{aligned} \sum \left( x^4 - 3x^3 +2x^2 + mx + n \right) &=& 0 \\ \sum x^4 - 3 \sum x^3 + 2 \sum x^2 + m \sum x + n \sum 1 & = & 0 \\ 9 - 3(3) + 2(5) + m(3) + n(4) & = & 0 \\ 3m + 4n &=& -10 \quad (\ast) \\ \end{aligned}

Now suppose that one of a , b , c a,b,c or d d is equals to 0 0 , WLOG a = 0 a = 0 . Then n = 0 m = b c d = 10 3 n = 0 \Rightarrow m =-bcd= -\frac{10}3 and by using the identity b 3 + c 3 + d 3 3 b c d = ( b + c + d ) ( ( b 2 + c 2 + d 2 ) ( b c + b d + c d ) ) b^3 + c^3 + d^3 - 3bcd = (b+c+d)( (b^2+ c^2+ d^2) - (bc+bd+cd)) . We get 3 3 ( 10 3 ) = 3 ( 5 2 ) 3 - 3\left( \frac{10}3\right) = 3\left(5 - 2\right) which is absurd. Thus a , b , c , d 0 a,b,c,d \ne 0 .

Thus for the equation x 4 3 x 3 + 2 x 2 + m x + n = 0 x^4 - 3x^3 +2x^2 + mx + n = 0 , the division by x x is possible.

( x 3 3 x 2 + 2 x + m + n x ) = 0 x 3 3 x 2 + 2 x + m 1 + n 1 x = 0 3 3 ( 5 ) + 2 ( 3 ) + m ( 4 ) + n 1 x = 0 4 m + n 1 x = 6 ( ) \begin{aligned} \sum \left( x^3 - 3x^2 +2x + m + \frac nx \right) &=& 0 \\ \sum x^3 - 3 \sum x^2 + 2 \sum x + m \sum 1 + n \sum \frac1x & = & 0 \\ 3 - 3(5) + 2(3) + m(4) + n \sum \frac1x & = & 0 \\ 4m + n \sum \frac1x & = & 6 \quad (\ast\ast) \\ \end{aligned}

Let f ( x ) = x 4 3 x 3 + 2 x 2 + m x + n f(x) = x^4 - 3x^3 +2x^2 + mx + n , then f ( x ) f(x) has roots a , b , c , d a,b,c,d . Similarly, x 4 f ( 1 x ) = 0 x^4 f\left(\frac1x\right) = 0 has roots 1 a , 1 b , 1 c , 1 d \frac1a, \frac1b,\frac1c,\frac1d .

x 4 f ( 1 x ) = n x 4 + m x 3 + 2 x 2 3 x + 1 1 a + 1 b + 1 c + 1 d = m n x^4 f\left(\frac1x\right) = nx^4 + mx^3 + 2x^2 -3x + 1 \Rightarrow \frac1a +\frac1b+\frac1c+\frac1d = -\frac mn

From ( ) (\ast \ast) , we get 4 m + n ( m n ) = 6 m = 2 4m + n\left( - \frac mn \right) = 6 \Rightarrow m = 2 thus from ( ) (\ast) , n = 4 n = -4 .

Hence the polynomial is x 4 3 x 3 + 2 x 2 + 2 x 4 x^4 - 3x^3 +2x^2 + 2x - 4 . With Rational Root Theorem, the two rational roots are 1 -1 and 2 2 by inspection. Factoring them out gives ( x + 1 ) ( x 2 ) ( x 2 2 x + 2 ) (x+1)(x-2)(x^2-2x+2) . So the other two roots are 1 ± i 1\pm i by quadratic formula.

WLOG, let a = 1 , b = 2 , c = 1 + i , d = 1 i a = -1, b = 2, c = 1 + i, d = 1- i . Then a 2015 = 1 a^{2015} = -1 . And by De Moivre's Theorem

c 2015 = ( 1 + i ) 2015 = ( 2 ) 2015 ( 1 2 + i 1 2 ) 2015 = 2 2015 / 2 ( cos ( π 4 ) + i sin ( π 4 ) ) 2015 = 2 2015 / 2 ( cos ( π 4 2015 ) + i sin ( π 4 2015 ) ) = 2 2015 / 2 ( cos ( π 4 ) + i sin ( π 4 ) ) \begin{aligned} c^{2015} &=& (1 + i)^{2015} \\ & =& (\sqrt2 )^{2015} \left( \frac1{\sqrt2} + i \frac1{\sqrt2} \right)^{2015} \\\ &=& 2^{2015/2} \left( \cos\left(\frac\pi4 \right) + i \sin\left(\frac\pi4 \right) \right)^{2015} \\ &=& 2^{2015/2} \left( \cos\left(\frac\pi4 \cdot 2015\right) + i \sin\left(\frac\pi4 \cdot 2015 \right) \right) \\ &=& 2^{2015/2} \left( \cos\left(\frac\pi4\right) + i \sin\left(\frac\pi4\right) \right) \\ \end{aligned}

Analogously, d 2015 = 2 2015 / 2 ( cos ( π 4 ) i sin ( π 4 ) ) d^{2015} = 2^{2015/2} \left( \cos\left(\frac\pi4\right) - i \sin\left(\frac\pi4\right) \right) . So c 2015 + d 2015 = 2 2015 / 2 2 cos ( π 4 ) = 2 2015 / 2 + 1 1 / 2 = 2 1008 c^{2015} + d^{2015} = 2^{2015/2} \cdot 2 \cos\left( \frac\pi4\right) = 2^{2015/2 + 1 - 1/2} = 2^{1008} .

Putting them all together yields 1 + 2 2015 + 2 1008 -1 + 2^{2015} + 2^{1008} . This suggests that p = 2 , q = 2015 , r = 1008 , s = 1 p = 2, q = 2015, r = 1008, s = 1 . However, we need to prove that p p is unique.

By Euclidean Algorithm, gcd ( 2015 , 1008 ) = gcd ( 1008 , 1007 ) = 1 \gcd(2015,1008) = \gcd(1008,1007) = 1 thus 2015 2015 and 1008 1008 don't share any common factor thus p p is indeed unique. Hence, our answer is p + q + r + s = 2 + 2015 + 1008 + 1 = 3026 p+q+r+s= 2 + 2015 + 1008 +1 = \boxed{3026} .

Moderator note:

Newton's sum would be a more direct way to obtain the coefficients of the polynomial.

Another approach would be to use Newton's sum to obtain the linear recurrence for T n = a n + b n + c n + d n T_n = a^n + b^n + c^n + d^n ,

good solution and new technique thanks for your answer :)

uzumaki nagato tenshou uzumaki - 5 years, 11 months ago

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No problem! I thought my technique will be shorter without Newton's Sum, turns out I'm so wrong. haha!

Pi Han Goh - 5 years, 11 months ago

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