Divisibility of polynomials?

Here's a simpler proof of #2:

n^n - n^2 + n - 1 = n(n-1)(n^(n-2)+n^(n-1)+...+1) - (n-1)^2

It suffices to prove P(n) = n^(n-2) + n^(n-1) + ... + 1 is divisible by n-1, which is true since

P(n) = [n^(n-2)-1] + [n^(n-1)-1] + ... + [1-1] + (n-1),

but n^k-1 = (n-1)(n^(k-1) + ... + 1), and we are done.

EDIT:

For #1 I don't understand why gianlino uses calculus to show the linear term is 0. By expanding with binomial theorem, it is easy to see the linear term coefficient is

2^(2n-1)(2n*b^(2n-1) - 2n(2b)^(2n-1), which is obvious 0.

Edit:

I've got it now, but if you are weak of heart don't look!

before we continue some restrictions are in oreder here:

if n = 0 then both a and b must be nonzero because a^(2*0) = a^0 and 0^0 has no sense!

if a = 0 we get b^(2n-2) * (2^(2n-1) - 1) which is integer only if n is nonzero

if b = 0 we get a^(2n-2) * (2^(2n-1) - 1) which is integer only if n is nonzero

So let's just assume that a and b are two different integers and n is a nonzero natural number.

Further on I'll use different algebra results:

Q is divisible by Q ---> (k*Q) is divisible by Q for any integer k

if X is divisible by Q and Y is divisible by Q the (X+Y) and (X-Y) are also divisible by Q

C(m,k) = k-combinations of m

C(m,k) = C(m,m-k)

C(m,k) = C(m,m) = 1 and C(m,1) = C(m,m-1) = m

Newton's binomial

(X + Y)^m = [C(m,0)* X^m] + [C(m,1)* X^(m-1) * Y] + [C(m,2)* X^(m-2) * Y^2] + [C(m,3)* X^(m-3) * Y^3] + ... + [C(m,m-2)* X^2 * Y^(m-2)] + [C(m,m-1)* X * Y^(m-1)] + [C(m,m)* Y^m]

As you can see all terms except the lsat two in Newton's binomial contain X^2 or a greater power of X. Because of that we'll factor X^2 from the terms that allow that and we'll get this:

(X + Y)^m = X^2 * [UglyTerm] + [C(m,m-1)* X * Y^(m-1)] + [C(m,m)* Y^m] ---> it won't matter how the "UglyTerm" looks like; all that matters is that it is integer if X and Y are integers.

(X + Y)^m = X^2 * [UglyTerm] + [m * X * Y^(m-1)] + [1 * Y^m]

(X + Y)^m = X^2 * [UglyTerm] + [m * X * Y^(m-1)] + Y^m ---> it won't matter how the "UglyTerm" looks like; all that matters is that it is integer if X and Y are integers.

On to the proof!

{[2^(2n-1)]*[a^(2n) + b^(2n)] - (a + b)^(2n) } / [(a-b)^2] is an integer actually means that:

{[2^(2n-1)]*[a^(2n) + b^(2n)] - (a + b)^(2n) } is divisible by [(a-b)^2]

For it to be more easyer drom now on I'll denote a-b = p.

a = p + b

a+b = a - b + 2b = p + 2b

Substituting these 3 in the relation above we get:

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{[2^(2n-1)]*[(p + b)^(2n) + b^(2n)] - (p + 2b)^(2n) } is divisible by p^2 ---> this is what we'll prove from now on!

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(p + b)^(2n) = p^2 * [UglyTerm1] + [C(2n,2n-1)* p * b^(2n-1)] + [C(2n,2n)* b^(2n)]

(p + b)^(2n) = p^2 * [UglyTerm1] + [(2n) * p * b^(2n-1)] + [(2n)* b^(2n)]

Since b and p are both integers so is "UglyTerm1"

(p + 2b)^(2n) = p^2 * [UglyTerm2] + [C(2n,2n-1)* p * (2b)^(2n-1)] + [C(2n,2n)* (2b)^(2n)]

(p + 2b)^(2n) = p^2 * [UglyTerm2] + [(2n) * p * (2b)^(2n-1)] + [1 * (2b)^(2n)]

(p + 2b)^(2n) = p^2 * [UglyTerm2] + [(2n) * p * 2^(2n-1) * b^(2n-1)] + [2^(2n) * b^(2n)]

Since b and p are both integers so is "UglyTerm2"

Substituting we get:

[2^(2n-1)]*[(p + b)^(2n) + b^(2n)] - (p + 2b)^(2n) =

= [2^(2n-1)]*{(p^2)*[UglyTerm1] + (2n)*p*[b^(2n-1)] + [b^(2n)] + b^(2n)} - {(p^2)*[UglyTerm2] + (2n)*p*[2^(2n-1)]*[b^(2n-1)] + [2^(2n)]*[b^(2n)]} =

= [2^(2n-1)]*{(p^2)*[UglyTerm1] + (2n)*p*[b^(2n-1)] + 2*[b^(2n)]} - {(p^2)*[UglyTerm2] + (2n)*p*[2^(2n-1)]*[b^(2n-1)] + [2^(2n)]*[b^(2n)]} =

= {2^(2n-1)*(p^2)*[UglyTerm1]} + {[2^(2n)]*(2n-1)*p*[b^(2n-1)]} + {2*[2^(2n-1)]*[b^(2n)]} - {(p^2)*[UglyTerm2]} - {(2n)*p*[2^(2n-1)]*[b^(2n-1)]} - {[2^(2n)]*[b^(2n)]} =

= {2^(2n-1)*(p^2)*[UglyTerm1]} + {[2^(2n)]*(2n-1)*p*[b^(2n-1)]} + {[2^(2n-1+1)]*[b^(2n)]} - {(p^2)*[UglyTerm2]} - {(2n)*p*[2^(2n-1)]*[b^(2n-1)]} - {[2^(2n)]*[b^(2n)]} =

= {2^(2n-1)*(p^2)*[UglyTerm1]} + {[2^(2n)]*(2n-1)*p*[b^(2n-1)]} + {[2^(2n)]*[b^(2n)]} - {(p^2)*[UglyTerm2]} - {(2n)*p*[2^(2n-1)]*[b^(2n-1)]} - {[2^(2n)]*[b^(2n)]} =

---> there are 6 terms above. The second and the fifth are identical except for the sign in front, this their sum is 0. The third and the sixth are identical except for the sign in front, this their sum is 0.

= {2^(2n-1)*(p^2)*[UglyTerm1]} - {(p^2)*[UglyTerm2]} =

= (p^2) * {2^(2n-1)*[UglyTerm1] - [UglyTerm2]} which is divisible by (p^2)

Follow closely the calculations and write them down if you must!

Best regards! And I hope you'll apreciate it!

For number 2 I made a terrible mistake, and the most elegant solution is provided by the answerer with chinese letters in his name.

For number 1 the proof I provided is the optimal and I put as much detail as I could.

2) Let P(n) = n^2 (n^(n-2) - 1) + (n-1) = (n-1) [ n^2( 1+n+....+n^(n-3)) + 1];

We just need to show that n-1 is a divisor of [ n^2( 1+n+....+n^(n-3)) + 1]

Working mod (n-1) we have n = 1. So

[ n^2( 1+n+....+n^(n-3)) + 1] = [1(1+1+.....+1) + 1] = [(1+ (n-3)) + 1] = (n-1) = 0

That's it.

1) Setting x = a - b, so that a = b + x and a + b = 2b + x, you need to show that

{[2^(2n-1)][(b+x)^(2n) + b^(2n)] - (2b+x)^(2n) } / x^2 is a polynomial with integer coefficients.

So you need to show that the constant and linear terms of

{[2^(2n-1)][(b+x)^(2n) + b^(2n)] - (2b+x)^(2n)} both vanish.

When x = 0 we have {[2^(2n-1)][(b)^(2n) + b^(2n)] - (2b)^(2n) = 2^(2n)b^(2n) - (2b)^(2n) = 0,

so the constant term vanishes. For the linear term we differentiate at 0. We get

{[2^(2n-1)][(b+x)^(2n) + b^(2n)] - (2b+x)^(2n)}'(0) = {[2^(2n-1)][2n(b)^(2n-1)] - 2n(2b)^(2n-1)} = 0

So {[2^(2n-1)][(b+x)^(2n) + b^(2n)] - (2b+x)^(2n) } / x^2 is indeed a polynomial and its coefficients are integers.

mathematical induction would be the best choice.