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.

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

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

\((n+1)^2−n^2=2n+1.\)
For sufficiently large \(N, 2n+1>b \quad \forall \quad n>N\). Now let \(m\) be large enough so that \(n_m>N\). Thus we have

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.

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

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

@Arian Tashakkor
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Nice solution! Exploiting the lack of consistency in the difference between two squares. I thought that there couldn't be more than 2 terms in such a sequence. So I tried to disprove it for 3 terms. Wrong approach.

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TopNewestIf \(a,b∈N\) then:

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

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

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

\((n+1)^2−n^2=2n+1.\) For sufficiently large \(N, 2n+1>b \quad \forall \quad n>N\). Now let \(m\) be large enough so that \(n_m>N\). Thus we have

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

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.

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

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Do you want an example of a progress with 3 terms?\(1^2,5^2,7^2\) Here you go.But Fermat proved that such progression cannot extend over 3 terms.

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

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

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Comment deleted May 15, 2015

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@Vishnu C sorry there was a typo.I edited it now

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