+73 1. Find the number of ordered pairs (a,b) of integers such that a+2a+5=b3
2. Find all complex numbers z such that z4=−4
Note: All solutions should be expressed in the form a+bi, where a and b are real numbers.
1. There is only one solution: (a,b) = (4,2).
2. By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4). Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i. Then the other roots work out as
4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,
4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and
4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.
+26396 1.
Find the number of ordered pairs (a,b) of integers such that a+2a+5=b3
a+2a+5=b33(a+2)=b(a+5)3a+6=ab+5b−3a+5b+ab=6−3a+5b+ab+(−3∗5)=(−3∗5)+6−3a+5b+ab−15=−15+6(a+5)(b−3)=−9m∗n=−9find all integer products m∗n=−9,where m=a+5, and n=b−3
all integer products
m∗nmna=m−5b=n+3−9=1×−91−9−4−6−9=−1×9−19−612−9=3×−33−3−20−9=−3×3−33−86−9=9×−19−142−9=−9×1−91−144
There are 6 integer products.
The number of ordered pairs (a,b) are 6.
(−4,−6), (−6,12), (−2,0), (−8,6), (4,2), (−14,4)
+73 oh wait no its right, im so sorry. btw i really liked the way you explained it, i understand now! :)
