⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ 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
Given that a , b , c and d are complex numbers that satisfy the equation above. If a 2 0 1 5 + b 2 0 1 5 + c 2 0 1 5 + d 2 0 1 5 can be written as p q + p r − 1 for positive integers p , q , r , evaluate p + q + r + 1 .
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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 is unique.
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Well, ( 1 0 0 8 , 2 0 1 5 ) = 1 so it is unique.
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Note that if the value of s is not stated, then we could have multiple solutions that depend on s .
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@Calvin Lin – What is the other solution for s another s = 1 sir?
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@Uzumaki Nagato Tenshou Uzumaki – For example, we could have p = 3 , q = 3 0 0 0 , r = 3 and s = 3 3 0 0 0 + 3 3 − 2 2 0 1 5 − 2 1 0 0 8 . 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 . There is still the work of showing that we have a unique answer. As pointed out, this is equivalent to g cd ( 1 0 0 8 , 2 0 1 5 ) = 1 .
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@Calvin Lin – Thanks you sir i haven't think so far. And thanks too for corect that :)
same with my solution . good but u have to show that ( 2 1 + 2 1 i ) 2 0 1 5 = 2 2 0 1 5 2 1 0 0 7 ( 2 1 − 2 1 i ) from De' Moivre or we can use ( 1 + i ) 4 = − 4
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Hello, I have edited my solution. Is it clear now?
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 ) , upon substitution, we get a b + a c + a d + b c + b d + c d = 2 3 2 − 5 = 2 .
Since the values of a , b , c and d satisfy these 4 equations and are interexchangeable, then there exist a 4th degree polynomial that have roots a , b , c and d . By Vieta's formula, the polynomial must be
x 4 − 3 x 3 + 2 x 2 + m x + n
for unique constants m and n . Take its sum:
∑ ( x 4 − 3 x 3 + 2 x 2 + m x + n ) ∑ x 4 − 3 ∑ x 3 + 2 ∑ x 2 + m ∑ x + n ∑ 1 9 − 3 ( 3 ) + 2 ( 5 ) + m ( 3 ) + n ( 4 ) 3 m + 4 n = = = = 0 0 0 − 1 0 ( ∗ )
Now suppose that one of a , b , c or d is equals to 0 , WLOG a = 0 . Then n = 0 ⇒ m = − b c d = − 3 1 0 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 ) ) . We get 3 − 3 ( 3 1 0 ) = 3 ( 5 − 2 ) which is absurd. Thus a , b , c , d = 0 .
Thus for the equation x 4 − 3 x 3 + 2 x 2 + m x + n = 0 , the division by x is possible.
∑ ( x 3 − 3 x 2 + 2 x + m + x n ) ∑ x 3 − 3 ∑ x 2 + 2 ∑ x + m ∑ 1 + n ∑ x 1 3 − 3 ( 5 ) + 2 ( 3 ) + m ( 4 ) + n ∑ x 1 4 m + n ∑ x 1 = = = = 0 0 0 6 ( ∗ ∗ )
Let f ( x ) = x 4 − 3 x 3 + 2 x 2 + m x + n , then f ( x ) has roots a , b , c , d . Similarly, x 4 f ( x 1 ) = 0 has roots a 1 , b 1 , c 1 , d 1 .
x 4 f ( x 1 ) = n x 4 + m x 3 + 2 x 2 − 3 x + 1 ⇒ a 1 + b 1 + c 1 + d 1 = − n m
From ( ∗ ∗ ) , we get 4 m + n ( − n m ) = 6 ⇒ m = 2 thus from ( ∗ ) , n = − 4 .
Hence the polynomial is x 4 − 3 x 3 + 2 x 2 + 2 x − 4 . With Rational Root Theorem, the two rational roots are − 1 and 2 by inspection. Factoring them out gives ( x + 1 ) ( x − 2 ) ( x 2 − 2 x + 2 ) . So the other two roots are 1 ± i by quadratic formula.
WLOG, let a = − 1 , b = 2 , c = 1 + i , d = 1 − i . Then a 2 0 1 5 = − 1 . And by De Moivre's Theorem
c 2 0 1 5 = = = = = ( 1 + i ) 2 0 1 5 ( 2 ) 2 0 1 5 ( 2 1 + i 2 1 ) 2 0 1 5 2 2 0 1 5 / 2 ( cos ( 4 π ) + i sin ( 4 π ) ) 2 0 1 5 2 2 0 1 5 / 2 ( cos ( 4 π ⋅ 2 0 1 5 ) + i sin ( 4 π ⋅ 2 0 1 5 ) ) 2 2 0 1 5 / 2 ( cos ( 4 π ) + i sin ( 4 π ) )
Analogously, d 2 0 1 5 = 2 2 0 1 5 / 2 ( cos ( 4 π ) − i sin ( 4 π ) ) . So c 2 0 1 5 + d 2 0 1 5 = 2 2 0 1 5 / 2 ⋅ 2 cos ( 4 π ) = 2 2 0 1 5 / 2 + 1 − 1 / 2 = 2 1 0 0 8 .
Putting them all together yields − 1 + 2 2 0 1 5 + 2 1 0 0 8 . This suggests that p = 2 , q = 2 0 1 5 , r = 1 0 0 8 , s = 1 . However, we need to prove that p is unique.
By Euclidean Algorithm, g cd ( 2 0 1 5 , 1 0 0 8 ) = g cd ( 1 0 0 8 , 1 0 0 7 ) = 1 thus 2 0 1 5 and 1 0 0 8 don't share any common factor thus p is indeed unique. Hence, our answer is p + q + r + s = 2 + 2 0 1 5 + 1 0 0 8 + 1 = 3 0 2 6 .
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 ,
good solution and new technique thanks for your answer :)
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No problem! I thought my technique will be shorter without Newton's Sum, turns out I'm so wrong. haha!
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Let 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 .
Define 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 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 Solving the system, we get ( k 3 , k 2 , k 1 , k 0 ) = ( − 3 , 2 , 2 , − 4 ) Thus, the polynomial with roots a , b , c , d is x 4 − 3 x 3 + 2 x 2 + 2 x − 4
Applying the Rational Root Theorem, we find that 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 ) Then we apply the quadratic equation and find that the remaining two roots are 1 ± i .
So the roots are ( a , b , c , d ) = ( − 1 , 2 , 1 + i , 1 − i ) . Now we can directly compute P 2 0 1 5 :
P 2 0 1 5 = ( − 1 ) 2 0 1 5 + 2 2 0 1 5 + ( 1 + i ) 2 0 1 5 + ( 1 − i ) 2 0 1 5 = − 1 + 2 2 0 1 5 + 2 2 0 1 5 ( 2 1 + 2 1 i ) 2 0 1 5 + 2 2 0 1 5 ( 2 1 − 2 1 i ) 2 0 1 5 = − 1 + 2 2 0 1 5 + 2 2 0 1 5 cis ( 4 π ) 2 0 1 5 + 2 2 0 1 5 cis ( − 4 π ) 2 0 1 5 = − 1 + 2 2 0 1 5 + 2 2 0 1 5 cis ( 4 2 0 1 5 π ) + 2 2 0 1 5 cis ( − 4 2 0 1 5 π ) = − 1 + 2 2 0 1 5 + 2 2 0 1 5 cis ( − 4 π ) + 2 2 0 1 5 cis ( 4 π ) = − 1 + 2 2 0 1 5 + 2 1 0 0 7 ⋅ 2 ( 2 1 − 2 1 i ) + 2 1 0 0 7 ⋅ 2 ( 2 1 + 2 1 i ) = − 1 + 2 2 0 1 5 + 2 1 0 0 7 ⋅ ( 1 − i ) + 2 1 0 0 7 ⋅ ( 1 + i ) = 2 2 0 1 5 + 2 1 0 0 8 − 1 So ( p , q , r , s ) = ( 2 , 2 0 1 5 , 1 0 0 8 , 1 ) and our final answer is 2 + 2 0 1 5 + 1 0 0 8 + 1 = 3 0 2 6