Can you prove this one too?

Show that there can never be an infinite arithmetic progression whose terms are made of distinct integers all of which are squares. Also, if you can, find the maximum size (number of terms) of such a progression.

#NumberTheory

Note by Vishnu C
6 years, 1 month ago

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Comments

If a,b∈Na,b∈N then:

f(m)=a+mbf(m)=a+mb

is an increasing function of m. Moreover, f(m+1)−f(m)=bf(m+1)-f(m)=b.

Suppose that f(m)=nm2f(m)=n^2_m for a sequence of terms nm∈Nn_m∈N. Necessarily nm is an increasing sequence of integers. We know that the gap between two consecutive squares is given by:

(n+1)2−n2=2n+1.(n+1)^2-n^2=2n+1. For sufficiently large N,2n+1>b∀n>NN, 2n+1>b \quad \forall \quad n>N. Now let mm be large enough so that nm>Nn_m>N. Thus we have

f(m+1)−f(m)=(nm+1)2−(nm)2≥(nm+1)2−(nm)2=2nm+1>bf(m+1)-f(m)=(n_{m+1})^2-(n_m)^2≥(n_m+1)^2-(n_m)^2=2n_m+1>b

Fermat has proven that there cannot exist such a progression beyond 3 terms and I didn't make head or tails of his proof.Just in case anyone was interested.

Arian Tashakkor - 6 years, 1 month ago

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I made a guess that it would be 3. But I couldn't quite get there. I tried to prove that you can't have a progression with 3 terms like this:

Say l2l^2 is the middle term and (l−k)2(l-k)^2 is the term below it. So now the common diference is 2kl−k22kl-k^2. Say now that (l+m)2(l+m)^2 is the next term. So the common difference can now be expressed as 2lm+m22lm+m^2.

Equating the common differences, we get that l=k2+m22(k−m)l=\frac{k^2+m^2}{2(k-m)}. And then I kinda got stuck

vishnu c - 6 years, 1 month ago

Did Fermat prove that it cannot exist with 3 terms as well? Or is there a progression with 3 terms?

vishnu c - 6 years, 1 month ago

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Do you want an example of a progress with 3 terms?12,52,721^2,5^2,7^2 Here you go.But Fermat proved that such progression cannot extend over 3 terms.

Arian Tashakkor - 6 years, 1 month ago
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