Recent content by rbetan

Okey let me recap and show what i think is some progress. Assume: n|a^n-b^n then a^n-b^n=cn for some integer c. but a =/=b therefore, a-b=/=0. so one can divide by a-b. (a^n-b^n)/(a-b) = (cn)/(a-b); "new goal is to show that n|cn/(a-b);" However, recall that...

i see how to get the next X from the previous (x,y) from your hint. if (x+y)/2= odd# ; keep it as the next X value and solve for the corresponding Y.if (x+y)/2= even#; then do |x-y|/2=odd#; keep this odd# as the next X value, solve for the corresponding Y. It is also clear from this process...

hey gunch: I do not see how does it follow that 11|(n+7)(n-4)+33; can anyone explain this step. Now taking for granted that 11|(n+7)(n-4)+33. then (n+7)(n-4)+33= 0(mod 11) so (n+7)(n-4)=-33(mod11); but -33(mod11)=0; so (n+7)(n-4)=0(mod11) so it follows that 11|(n+7)(n-7)...

Homework Statement prove that 11^2 does not divide n^2+3n+5; for any n. In order for this to make sense n must be an integer. Homework Equations The Attempt at a Solution want to show that n^2+3n+5 is not congruent to 0(mod 121) Assume towards a contradiction that 11^2...

I do not see how does it follow that c/(a-b) must be an integer. can you please help.

thanks for your help, i really appreciate it

if one can show that n | cn/(a-b) then it is done. n| cn/(a-b) means that cn/(a-b)=nK; where K is an integer. so for n to divide cn/(a-b) means that c/(a-b) is an integer But how can i know that indeed this quantity c/(a-b) is an integer?

then (a^n-b^n)/(a-b) is equal to (cn)/(a-b)

I am sorry i did not understand what you meant in this last post. can you elaborate more please. i did understand that since a is different from b then their difference a-b is different from zero. But i did not understand what do you mean by replacing a^n-b^n with cn.

okey i got an idea from your hint, but i get stuck again, let me show you what i got so far: assuming that a^n+b^n=cn for some c but one can factor a^n+b^n=(a-b)(a^p-1+a^n-2 b+…+a b^n-2+b^n-1) so (a^n+b^n)/(a-b)=(a^p-1+a^n-2 b+…+a b^n-2+b^n-1) if one can show that n|(a^p-1+a^n-2 b+…+a...

Here are some of the numerical examples... the higher the n the longer it took me to spot a solution. Numerical Examples: I could not get 2^1,2^2, but i did get the following 2^3=7(1)^2+(1)^2 so for n=3; one solution is x=1, y=1 2^4=7(1)^2+(3)^2 so for n=4; one solution is x=1, y=3...

[b]1. Homework Statement . show that if n devides a^n-b^n then n devides the quotient (a^n-b^n)/(a-b). here n,a,b are natural numbers with a and b distinct. [b]2. Homework Equations . [b]3. The Attempt at a Solution . i know i have to assume that n divides a^n-b^n. So, then...

Homework Statement Show that any power of two can be represented as 2^n=7*x^2+y^2 for some x,y odd. This problem was given to us in preparation for the final for an undergraduate number theory class. Homework Equations The Attempt at a Solution I tried numerical example to...

Describe All n such that 3 divides n times 2 raised to the n, plus one.( n*2^n + 1) I know that a number is divisible by 3 if the sum of its digits adds up to a number that itself is divisible by 3. But this is probably not helpful for this problem. I also know that a number A is...