[Peter]

Suppose that and are distinct real numbers such that are all integers. Show that and are integers.

[edriv]

First, we prove that a,b are rational numbers.





Then we can find three integers a',b',n such that:



The hypotesis translates to .
Now, and we can suppose . Therefore .

Let d be the integer such that . First, we suppose d>1.
We have . We get , but , therefore and . (in this step we must remember that d>1).

Now we consider 2(d+1). We know that . But , then and . But , then finally .
But therefore, n=1 and a,b are integers.

If , then and and , 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 . Let us take a positive integer k such that (this obviously exists).
But then . We must find factors 2 in this product. But if x,y are odd integers, . And by hypotesis, and by hypotesis. Therefore we find exactly factors 2 in this product, which is too low.

I hope it is correct!

I hope it is correct!

[bambaman]

Lets say we showed that a,b are rational (like edriv did). Let:

. If we show that , we get: , , or: are integers. It follows from the following lemma:

Lemma: If for all , where , are integers, then: .
Proof: Let p be a prime divisor of a, , (). We have, by "Lifting the Exponent" lemma:




or (in the case p=2):
.

We can put that together in the following way - .

It implies that: for all n>1. Let . We have: , which implies that:
for all n>1. It can't happen for large n unless or .

First case - . It means that , or: for all n, which gives , and then also .
Second case - , or: .

Q.E.D.

"Lifting the Exponent": If p is an odd prime, and a,b are integers, and: , we have:
If p=2, , we have:

I have dropped out some minor details, ask a question or point an error if you find one.

[No Reason]

Sorry for spam.But what "GML" mean ?

[hsiljak]

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

[xxp2000]

Here is a proof a little bit different from edriv. It is the same up to
, .

We can also assume wlog.

Let , we know
, where is binomial coefficients.
We can always find big with .
Note the first term on RHS is multiple of . So we have , or .
Then the first term on RHS is multiple of , or . So we have , or .
Then the first term on RHS is multiple of , or . So we have , or .
....
We have for any big integer , being nonzero finite number implies . Hence and