A few days ago
Electra

# modern math ?s?

doing any one of these will help alot..thank you!

1) Let A and B intergers. Prove: If is a factor of B, then A^2 is a factor of B^2.

2) Let A and B be natural numbers such that A^2=B^3 Prove: If 4 divides B, then 8 divides A.

3) Prove that for a1 intergers A and M, if A and M are the lengths of sides of a right triangle and m+1 is the length of the hypotenuse, then A is an odd interger.

4) Prove: if A and B are odd intergers, the 4 doesnt divide A^2 + B^2.

A few days ago
Kia

1)

If A is a factor of B, then we can write B as A*C for some integer C.

Then B^2 = (A*C)^2 = A^2 * C^2

Since A^2 multiplied by some positive integer results in B^2, therefore A^2 is a factor of B^2

2)

If 4 divides B, the we can write B = 4 * C for some natural number C.

Then A^2 = B^3 = (4 * C)^3 = 4^3 * C^3

A = sqrt(4^3 * C^3)

A = 4C sqrt(4*C)

A = 4C sqrt(2^2 * C)

A = 8C sqrt(C)

Since 8 is a factor of A, therefore 8 divides A

3)

In a right triangle a^2 + b^2 = c^2

If A and M are the lengths of sides of a right triangle and m+1 is the length of the hypotenuse, then:

A^2 + M^2 = (M + 1)^2

A^2 + M^2 = M^2 + 2M + 1

A^2 = 2M + 1

For any integer M, 2M is always an even number and 2M + 1 is always an odd number. In here, A^2 is an odd number, therefore A must be an odd number, since the square of any even number is even.

4)

if A and B are odd integers, then we can write

A = 2m + 1 (for some integer m)

B = 2n + 1 (for some integer n)

Then

A^2 + B^2

= (2m + 1)^2 + (2n + 1)^2

= (4m^2 + 4m + 1) + (4n^2 + 4n + 1)

= 4m^2 + 4m + 1 + 4n^2 + 4n + 1

= 4m^2 + 4n^2 + 4m + 4n + 1 + 1

= 4m^2 + 4n^2 + 4m + 4n + 2

Since we cannot factor 4 from the equation above, therefore 4 does not divide A^2 + B^2

I hope this helped (even though a bit late perhaps)

Kia

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