Hi I'm stuck on part 1 of this question.

I have attached the question.

Could someone please tell me what to do.

Part 1 being the bit about x^2+5y=243723

Thanks

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- Feb 23rd 2013, 10:49 AMJames7361539Can someone please help with this Sequences Q
Hi I'm stuck on part 1 of this question.

I have attached the question.

Could someone please tell me what to do.

Part 1 being the bit about x^2+5y=243723

Thanks - Feb 23rd 2013, 11:18 AMOptikalRe: Can someone please help with this Sequences Q
I would prove this by considering the equation modulo 5. In this case, the 5y term will disappear because 5 is congruent to 0 mod 5. Thus, it suffices to show that

x^2 cannot be congruent to 3 mod 5 (since 243723 is congruent to 3 mod 5).

To do this, consider all the possibilities for x. We can have x=0,1,2,3,4 and so x^2=0,1,4,4,1 mod 5 respectively.

Thus, x^2 cannot be congruent to 3 mod 5 and so we are done.

Hope this helps!

Optikal - Feb 23rd 2013, 11:21 AMHallsofIvyRe: Can someone please help with this Sequences Q
Do you understand the connection between this problem and the first paragraph? Every integer can be written in one of forms:

5n, 5n+ 1, 5n+ 2, 5n+ 3, or 5n+ 4- in other words, every integer has remainder 0 to 4 when divided by 5.

Suppose x is a mutiple of 5: x= 5n. Then x^2+ 5y= 25n^2+ 5y= 5(5n^2+ y), a multiple of 5. But 243723 is NOT multiple of 5.

(5n+ 1)^2= 25n^2+ 10n+ 1= 5(5n^2+ 2n)+ 1 has remainder 1 when divided by 5.

(5n+ 2)^2= 25n^2+ 20n+ 4= 5(5n^2+ 4n)+ 4 has remainder 4 when divided by 5.

What about the others? What are the possible remainders, when divided by 5, of ? - Feb 24th 2013, 03:08 AMJames7361539Re: Can someone please help with this Sequences Q
Hey I am starting to see it but I still don't understand.

First of all I have written out the five sequences in the form Un=a+(n-1)d

I don't know, maybe you do arithmetic sequences different in America? Un just means the nth term.

Therefore I have

For A: Un=1+(n-1)5

For B: Un=2+(n-1)5

For C: Un=3+(n-1)5

For D Un=4+(n-1)5

For E: Un=5+(n-1)5

I just don't understand how you have gotten your sequences?

Can you do it with my sequences?

Having reviewed your sequences I think I see that you are simply doing n>=0

whereas I am starting at n=1. IE n>=1

But I would still like to know if you can do it with my sequences?

I am currently working on it.

Hey I just did it with my sequences!

Thanks for your great explanation!(Happy)

I may need help for part 2 but I'm currently working on it and I may manage it myself. - Feb 24th 2013, 08:02 AMJames7361539Re: Can someone please help with this Sequences Q
Hey Halls of Ivy

I am kind of stuck on part 2

I think it may be something about considering x^4 mod 2 and showing there is always a remainder? Am I right?

Anyways I let x = 5+(n-1)5

and x^4= 625(n-1)^4+2500(n-1)^3+3750(n-1)^2+2500(n-1)+625

Anyways if I'm right how do I show that this is not divisible by two?