Algebra 2 Questions

[rao_anandini]

1) x^4-11x^2+1

2) If M is a multiple of p, and N is a multiple of p, prove that M+N is also a multiple of p.

3) Find the general integer solutions to the equation: 5x-8y=1

4) Prove that 2 consecutive integers n, and n+1 have no common prime factors.

 

[jwpaine]

Those are some nice questions. What have you tried on your own?

John.

 

[tkhunny]

#1 looks awfully quadratic in form. Surely you can deal with that one.

 

[morson]

1) Don't know what you want for this one.

2) If M is a multiple of p, then by definition, M = kp, where k is an integer.

Similarly, if N is a multiple of p, then N = tp, where t is an integer.

So \(\displaystyle \L\ M + N = kp + tp = (k + t)p\)

3) There's a way of finding the integer solutions in terms of a parameter t. Euclid and Diophantine invented it centuries ago. In general, if \(\displaystyle \L\ (x_0, y_0)\) was an integer solution for:

\(\displaystyle \L\ ax + by = c\)

Then: \(\displaystyle \L\ x = x_0 + t\frac{b}{d}\\) , \(\displaystyle y = y_0 - t\frac{a}{d}\\) , \(\displaystyle t \in\ J\)

where d is the highest common factor of a, b and c.

For your problem, \(\displaystyle \L\ (x_0, y_0) = (5, 3)\)

4) Assume they did.

 

[TchrWill]

3) Find the general integer solutions to the equation: 5x-8y=1

5x - 8y = 1
Dividing through by 5 yields x - y - 3y/5 = 1/5
(3y + 1)/5 must be an integer as does (6y + 2)/5
Dividing by 5 again yields y + y/5 + 2/5.
(y += 2)/5 must be an integer k making y = 5x - 2
Substituting back into 5x - 8y = 1 yields x = 8k - 3.
For positive integer solutions, k can be 1 or higher.
k...1...2...3...4...5...6...
x...5..13..21..29..37..45 and so on
y...3...8..13..18..23..28 and so on.

Of course, negative solutions are present with k's equal to 0 or less.