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?

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%26gt;= 0, -x if x %26lt; 0}. prove: for every a,b, element of R, |a+b|%26lt;=|a|+|b|.

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?
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 =%26gt; a|(a+b)^2.





b = 0 mod a =%26gt; b+a = 0 mod a =%26gt; (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%26lt;=|x|, and -x%26lt;=|x|. If a+b %26gt;= 0, then |a+b|=a+b. Since a%26lt;=|a| and |b|%26lt;=|b|, then a+b%26lt;=|a|+|b|, so it holds. If a+b%26lt;0, then |a+b|=-a-b, but -a%26lt;=|a| and -b%26lt;=|b|, so again it holds.