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About LinkBacksHi 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
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
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?
Hey I am starting to see it but I still don't understand.Originally Posted by HallsofIvy
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
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!
I may need help for part 2 but I'm currently working on it and I may manage it myself.
Hey Halls of IvyDo 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
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?
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