• February 24th 2013, 11:30 AM
James7361539
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.
• February 24th 2013, 11:50 AM
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.
• February 24th 2013, 12:05 PM
James7361539
Quote:

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?
• February 24th 2013, 12:34 PM
emakarov
Quote:

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".
• February 24th 2013, 12:59 PM
James7361539
Quote:

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?
• February 24th 2013, 01:27 PM
emakarov