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We have:
n3 - 8n2 + 20n - 13 = (n - 1)(n2 - 7n +13)
Because n3 - 8n2 + 20n - 13 are prime numbers
2
1 1
7 13 is
<i>n</i>
<i>SO</i>
<i>n</i> <i>n</i> <i>a prime</i>
2
1 is
or
7 13 1
<i>n</i> <i>a prime</i>
<i>n</i> <i>n</i>
<b>Problem 1</b>: How many positive integers n are
A= prime numbers
2
If a, b, c are real numbers so that a2 + 4b = 7;
b2 +8c = -10 and c2+ 6a = -26. Find T = a2+ b3+ c4.
2
2 2 2 2
2
4 7
8 10 a + 4b+b +8c+c +6a = 7+(-10)+(-26)
6 26
<i>a</i> <i>b</i>
<i>We have</i> <i>b</i> <i>c</i>
<i>c</i> <i>a</i>
a2+ 4b + b2+ 8c + c2+ 6a + 29 = 0
(a + 3)2+ (b + 2)2+ (c + 4)2= 0
2
3
4
2 3 4
( a2+ 6a + 9) + ( b2+ 4b + 4) + (c2 + 8c + 16) = 0
<b>Problem 3:</b> Find the balance polynomial divided by
polynomial P(x) =5 + x + x3 + x9 + x27 + x81 for
polynomial Q(x) = x2 - 1
We have:
= x(x2 - 1) + x(x8 - 1) + x(x26- 1)+x(x80 - 1) + 5x + 5
Note that a2n – b2n(a - b) from n<sub></sub>N.So (x2n-1)<sub></sub>(x2 - 1)
P(x) : Q(x) balance polynomial 5x + 5.
Therefore balance polynomial divided by
polynomial P(x) for polynomial Q(x) is 5x +5.
Let balance polynomial divided by polynomial P(x) for
polynomial Q(x) is R(x) = ax + b (a; b R)
We have: P(x) = (x2 - 1). A(x) + ax + b
(A(x) is quotient polynomial).
<i>Apply the Bezout the em</i>
<i>We have</i>
<i>P</i> <i>a b</i>
<i>a</i> <i>b</i>
<i>P</i> <i>a b</i>
Find the balance polynomial divided by polynomial
P(x) = x81 + x49 + x25 + x9 +x + 1
- Review all the exercises that we do today
- Solve the following exercises:
<b>Question 1</b>: Find the numbers of different positive
integer triples (x; y; z) that satisfy equations
x2 + y - z = 100 and x + y2 - z = 124.
<b>Question 2</b>: Find the natural numbers x; y; z that
satisfy the following conditions:
x3 + y3 = 2z3 x + y + z is a prime number