N 17

Peter

N 17

**Posted:** Fri May 25, 2007 2:25 am

edriv

First, we prove that a,b are rational numbers.
$a-b \in \mathbb{Q}$
$a+b = \frac{a^2-b^2}{a-b} \in \mathbb{Q}$
$a = \frac{(a+b)+(a-b)}2 \in \mathbb{Q}$
$b = \frac{(a+b)-(a-b)}2 \in \mahtbb{Q}$

Then we can find three integers a',b',n such that:
$a = \frac{a'}n$
$b = \frac{b'}n$
$d \mid a' \land d \mid b' \land d \mid n \Rightarrow |d| = 1$
The hypotesis translates to $n^k \mid a'^k - b'^k \quad \forall k \in \mahtbb{N}$ .
Now, $n \mid a'-b'$ and we can suppose $(a',n) = 1$ . Therefore $(b',n) = 1$ .

Let d be the integer such that $n^d \mid \mid a'-b'$ . First, we suppose d>1.
We have $n^{d+1} \mid a'^{d+1} - b'^{d+1} = (a'-b')(a'^d + \ldots + b'^d)$ . We get $n \mid (a'^d + \ldots + b'^d)$ , but $\gcd(a'-b', a'^d + \ldots + b'^d) \mid n$ , therefore $n \mid \mid (a'^d + \ldots + b'^d)$ and $n^{d+1} \mid \mid a'^{d+1} - b'^{d+1}$ . (in this step we must remember that d>1).

Now we consider 2(d+1). We know that $n^{2d+2} \mid a'^{2d+2} - b'^{2d+2} = (a'^{d+1} + b'^{d+1})(a'^{d+1}-b'^{d+1})$ . But $n^{d+1} \mid \mid a'^{d+1} - b'^{d+1}$ , then $n^{d+1} \mid a'^{d+1} + b'^{d+1}$ and $n^{d+1} \mid (a'^{d+1} + b'^{d+1}) + (a'^{d+1} - b'^{d+1}) = 2a'^{d+1}$ . But $(a',n) = 1$ , then finally $n^{d+1} \mid 2$ .
But therefore, n=1 and a,b are integers.

If $n \mid \mid a'-b'$ , then $n^2 \mid (a'+b')(a'-b')$ and $n \mid a'+b'$ and $n \mid 2a'$ , as before. Then n=2 and a',b' are odd integers such that a'-b' is a multiple of 2 but not of 4. Let us take d such that $2^d \mid \mid a'+b'$ . Let us take a positive integer k such that $2^k - k > d$ (this obviously exists).
But then $2^{2^k} \mid a'^{2^k} - b'^{2^k} = (a'^{2^{k-1}} - b'^{2^{k-1}}) \cdots (a'^2+b'^2)(a'+b')(a'-b')$ . We must find $2^k$ factors 2 in this product. But if x,y are odd integers, $2 \mid \mid x^2+y^2$ . And $2 \mid \mid a'-b'$ by hypotesis, and $2^d \mid \mid a'+b'$ by hypotesis. Therefore we find exactly $d + k$ factors 2 in this product, which is too low.

I hope it is correct!

I hope it is correct!

**Posted:** Fri May 25, 2007 2:25 am

bambaman

#3

Solution using a useful lemma

**Posted:** Sun Oct 19, 2008 7:33 pm

No Reason

#4

Sorry for spam.But what "GML" mean ?

**Posted:** Sun Dec 07, 2008 2:28 pm

hsiljak

#5

I assume it's George T. Gilbert, Mark I. Krusemeyer, Loren C. Larson, The Wohascum County Problem Book, MAA.

**Posted:** Sun Dec 07, 2008 6:33 pm

_________________
Engineer outside [mathematician inside].

xxp2000 · Riemann Hypothesis Offline Joined: 17 Oct 2008 Posts: 258

#6

Here is a proof a little bit different from edriv. It is the same up to
$n|a'-b'$ , $(n,a')=(n,b')=1$ .

We can also assume $n>0$ wlog.

Let $a'=An+b'$ , we know
$n^k|a'^k-b'^k=\sum_{i=2}^k c_i(An)^ib'^{k-i}+kAnb'^{k-1}$ , where $c_i$ is binomial coefficients.
We can always find big $k$ with $(k,n)=1$ .
Note the first term on RHS is multiple of $(An)^2$ . So we have $n^2|kAnb'^{k-1}$ , or $n|A$ .
Then the first term on RHS is multiple of $(An)^2$ , or $n^4$ . So we have $n^4|kAnb'^{k-1}$ , or $n^3|A$ .
Then the first term on RHS is multiple of $(An)^2$ , or $n^8$ . So we have $n^8|kAnb'^{k-1}$ , or $n^7|A$ .
....
We have $n^m|A$ for any big integer $m$ , $A$ being nonzero finite number implies $n=1$ . Hence $a=a'$ and $b=b'$

**Posted:** Sun Dec 07, 2008 7:45 pm