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# help

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Find the only real number that can be expressed in the form (a + bi)^3 - 107i, where i^2 = -1, and a and b are positive integers.

Jan 30, 2019

#1
+5651
+2

$$(a+i b)^3 - 107i = (a^3-3 a b^2) + i(3 a^2 b - b^3-107)\\ \text{If this is to be real, then}\\ 3 a^2 b - b^3-107 = 0\\ (3a^2-b^2)b = 107$$

$$\text{the only factors of 107 are }1,~107 \text{ so either}\\ b=1,~(3a^2-b^2)=107 \text{ or }b=107,~(3a^2 - b^2)=1$$

$$\text{If the first}\\ (3a^2 - b^2) = 3a^2 - 1=107\\ 3a^2 = 108\\ a^2 = 36\\ a=\pm 6$$

$$\text{If the second}\\ (3a^2 - b^2) = 3a^2 - 107^2 = 1\\ 3a^2 = 1 + 107^2 = 11450\\ a^2 = \dfrac{11450}{3}\\ \text{and this doesn't lead to }a \text{ being an integer}$$

$$\text{So our solution is}\\ r = (6+i)\\ (6+i)^3-107i = 198$$

.
Jan 30, 2019

#1
+5651
+2

$$(a+i b)^3 - 107i = (a^3-3 a b^2) + i(3 a^2 b - b^3-107)\\ \text{If this is to be real, then}\\ 3 a^2 b - b^3-107 = 0\\ (3a^2-b^2)b = 107$$

$$\text{the only factors of 107 are }1,~107 \text{ so either}\\ b=1,~(3a^2-b^2)=107 \text{ or }b=107,~(3a^2 - b^2)=1$$

$$\text{If the first}\\ (3a^2 - b^2) = 3a^2 - 1=107\\ 3a^2 = 108\\ a^2 = 36\\ a=\pm 6$$

$$\text{If the second}\\ (3a^2 - b^2) = 3a^2 - 107^2 = 1\\ 3a^2 = 1 + 107^2 = 11450\\ a^2 = \dfrac{11450}{3}\\ \text{and this doesn't lead to }a \text{ being an integer}$$

$$\text{So our solution is}\\ r = (6+i)\\ (6+i)^3-107i = 198$$

Rom Jan 30, 2019
#2
+102372
0

Nicely done, Rom.....!!!!

CPhill  Jan 30, 2019