prove.if f(x)=ax^2+bx+c, then if the inputs go up by 1 the second set of differences will be constant =(2!)a?

Question

Also prove.
1. let a,b be elements of Z. Prove if a^2 + 2b^2 is congruent to 0(modn), then either a and b are both congruent to 0 module 3 or neither is congurent to 0 modulo 3.

2. let a,b,c,d be element of Z. prove: for a|b if is necessary that a|(a+b)^2.

3. let a,b,c,d be element of Z. prove: if a is congruent to b(modn) and a is congurent to c(modn), then a is congurent to (2b-c)(modn).

4. definition: for a real number x, |x| = {x if x>= 0, -x if x < 0}. prove: for every a,b, element of R, |a+b|<=|a|+|b|.

Answer 1

f(x+1)=a(x+1)^2 + b(x+1) + c =
a(x^2+2x+1) + b(x+1) + c =
f(x) + 2ax + b
I don't have a definition for "second set of differences" but since you must have one, I'm sure you can complete this problem.

1. I assume you meant a^2 + 2b^2 is congruent to 0 (mod 3). Assume that one of a or b is congruent to 0 (mod 3), but not the other. For instance, a = k mod 3, k=1 or 2, and b = 0 mod 3. Then a^2 = 1 mod 3, and 2b^2 = 0 mod 3. Therefore a^2 + 2b^2 = 1 mod 3. Similar for if a = 0 mod 3 and b = k mod 3. By the contrapositive, the statement is proved.

2. You're proving that a|b => a|(a+b)^2.

b = 0 mod a => b+a = 0 mod a => (b+a)^2 = 0 mod a. To see this, suppose a|b, then b=ak for some integer k. Then b+a=a(k+1), and (b+a)^2 = a*(a(k+1)).

3. This appears to be typed wrong. How can a = b mod n and c mod n, unless b=c? Also what role does d play? Please check.

4. The key is that x<=|x|, and -x<=|x|. If a+b >= 0, then |a+b|=a+b. Since a<=|a| and |b|<=|b|, then a+b<=|a|+|b|, so it holds. If a+b<0, then |a+b|=-a-b, but -a<=|a| and -b<=|b|, so again it holds.