# Thread: Help needed for this Q about sequences please

1. ## Help needed for this Q about sequences please

Hey I have almost done this question but I'm stuck on part 2.
Could someone please point me in the right direction.
I have attached the question.
Thanks in advance for replies.

2. ## Re: Help needed for this Q about sequences please

Use a similar approach to part (i). Find to which sequences A - E belong $x^4$, $2y^4$ and $x^4 + 2y^4$ for all x and y and see if the right-hand side number belongs to those sequences.

3. ## Re: Help needed for this Q about sequences please

Originally Posted by emakarov
Use a similar approach to part (i). Find to which sequences A - E belong $x^4$, $2y^4$ and $x^4 + 2y^4$ for all x and y and see if the right-hand side number belongs to those sequences.
I'm sorry I don't understand what you mean by find to which sequences A - E belong etc etc.
I am trying to use a similar approach to part 1 but I just can't think on what to do?

4. ## Re: Help needed for this Q about sequences please

Originally Posted by James7361539
I'm sorry I don't understand what you mean by find to which sequences A - E belong etc etc.
The introductory part of the problem asked to you determine to which sequence belongs the square of every term in B and the square of every term in C. Also, recall HallsofIvy's post, which shows to which sequences belong squares of numbers. Here I suggest determining to which sequences belongs the fourth powers of numbers.

Here is a start. Since all sequences A - E together cover all numbers, you need to consider five cases. The nth term of, say, sequence C has the form 3 + 5(n - 1). Let k denote n - 1; then the form is 3 + 5k. After some thought it becomes clear that (3 + 5k)(3 + 5k)(3 + 5k)(3 + 5k) = 3^4 + 5(...) where ... denotes some big expression. Also, 3^4 = 81 = 1 + 5 * 16. So, (3 + 5k)^4 = 1 + 5 * j for some j. It is easy but tedious to see that j is a nonnegative integer, so 1 + 5 * j = 1 + 5(j' - 1) where j' = j + 1 ≥ 1. Thus, (3 + 5k)^4 belongs to sequence B. In all this, it is only important to find the remainder of 3^4 when divided by 5; all other terms resulting from expansion of (3 + 5k)^4 are multiples of 5 and thus don't influence to which sequence (3 + 5k)^4 belongs.

Following this pattern, for each sequence and each element in that sequence determine to which sequence belongs the fourth power of that element. In other words, for each positive integer x, take the remainder when x is divided by 5 (it determines the sequence to which x belongs) and use it to determine the remainder when x^4 is divided by 5 (i.e., to which sequence x^4 belongs). Then do the same with 2y^4 for each y. Finally, do this for x^4 + 2y^4. Remember that all that matters is the remainder when divided by 5.

Edit: "remainder of 3 when divided by 5" -> "remainder of 3^4 when divided by 5".

5. ## Re: Help needed for this Q about sequences please

Originally Posted by emakarov
The introductory part of the problem asked to you determine to which sequence belongs the square of every term in B and the square of every term in C. Also, recall HallsofIvy's post, which shows to which sequences belong squares of numbers. Here I suggest determining to which sequences belongs the fourth powers of numbers.

Here is a start. Since all sequences A - E together cover all numbers, you need to consider five cases. The nth term of, say, sequence C has the form 3 + 5(n - 1). Let k denote n - 1; then the form is 3 + 5k. After some thought it becomes clear that (3 + 5k)(3 + 5k)(3 + 5k)(3 + 5k) = 3^4 + 5(...) where ... denotes some big expression. Also, 3^4 = 81 = 1 + 5 * 16. So, (3 + 5k)^4 = 1 + 5 * j for some j. It is easy but tedious to see that j is a nonnegative integer, so 1 + 5 * j = 1 + 5(j' - 1) where j' = j + 1 ≥ 1. Thus, (3 + 5k)^4 belongs to sequence B. In all this, it is only important to find the remainder of 3^4 when divided by 5; all other terms resulting from expansion of (3 + 5k)^4 are multiples of 5 and thus don't influence to which sequence (3 + 5k)^4 belongs.

Following this pattern, for each sequence and each element in that sequence determine to which sequence belongs the fourth power of that element. In other words, for each positive integer x, take the remainder when x is divided by 5 (it determines the sequence to which x belongs) and use it to determine the remainder when x^4 is divided by 5 (i.e., to which sequence x^4 belongs). Then do the same with 2y^4 for each y. Finally, do this for x^4 + 2y^4. Remember that all that matters is the remainder when divided by 5.

Edit: "remainder of 3 when divided by 5" -> "remainder of 3^4 when divided by 5".
Hey thanks for this explanation.
I can totally see how this works but I just stuck with the Un=3+(n-1)5
and then [3+(n-1)5]^4=81+5(.....)
I don't see how it is necessary to do the whole thing with j? I didn't get that bit?
I just thought that since n>=1 then the 5(....) bit can either be 0 or a multiple of 5.
Then 81 + 5(...) mod 5 = 1 and I thought that was all that was required?

6. ## Re: Help needed for this Q about sequences please

Originally Posted by James7361539
I don't see how it is necessary to do the whole thing with j? I didn't get that bit?
I just thought that since n>=1 then the 5(....) bit can either be 0 or a multiple of 5.
Then 81 + 5(...) mod 5 = 1 and I thought that was all that was required?
Basically, you are right. I am just saying that not every real number is an element of one of the sequences A - E, but only positive integers are. In the introduction to the problem, when we were asked specifically to show that (2 + 5(n - 1))^2 is in D, it was not enough to show that (2 + 5(n - 1))^2 = 4 + 5m for some real m. For example, 4 + 5(-3) does not belong to D and neither does 4 + 5(1/3). We had to show that m is a nonnegative integer. However, all numbers that we deal in this problems are integers, and the nonnegativity requirement is artificial and is only used because of sequences A - E. It is probably better to speak about remainders when divided by 5 from the start.

So yes, the goal is to find possible remainders of x^4 + 2y^4 when divided by 5 and not to worry about the quotients.