NMO problem

[willpower]

I need help with the following problem:
Prove that if an arithmetical progression of positive integers contains a square, then it contains infinitely many squares.
Hope to get a solution asap. Thank you!

[sourav das]

Let $a_i = a_0 + n r $ is an arithmatic progression for $n= 0, 1, 2...$
Let $a_c=m^2 = a_0 + c r$ a perfect square in the sequence. then put $ n=c+2m+r $
then$a_{c+2m+r} = a_0 + (c+2m+r)r = a_0 +cr + 2mr + r^2 = m^2 +2mr + r^2=(m+r)^2$ And so there are infinitely many squares in the sequence.

[willpower]

Why do we put this expression "n=c+2m+r" into the original formula?
Also, I knew that the nth term of an arithmetic progression could be written as "a + (n-1)d". I'm not aware of the form used. :S

[*Mahi*]

Just replace $n-1 \rightarrow n$,$a \rightarrow a_0$ and $d \rightarrow r$ to derive one formula from another.
And the theme of this solution is called "completing the square"

[sourav das]

I just work backward.I assume if $a_c=m^2$ then let $a_x=k^2$ and $K>m$. We have to find $x$.Now $ K^2 -m^2=r(x-c)$. Now i just test it by putting $k-m=r$ then i found that $x=c+2m+r$ and it works.

[willpower]

I didn't get the last step. The part where you say: "Now i just test it by putting k−m=r then i found that x=c+2m+r and it works. " How come?
It's my first time at such questions. Kindly elaborate. Thanks.

[sourav das]

First, don't be afraid to question where you can't understand in this forum. In our L.H.S=$k^2 -m^2$ = $(k+m)(k-m)$
R.H.S.=$r(x-c)$ . So $r|(k+m)(k-m)$ . That's why i was curious what will happen if $r=k-m$. And the trick worked!

[willpower]

As far as I have gathered from the posts above, we first write down the general formula of an A.P:
$a_i=a_0+nr$
Then we represent a perfect square using this notation: $a_c = a_0+cr = m^2$
Let $a_x = k^2$ where $k > m$
=>$ k^2 - m^2 = a_x - a_c$
=> $(k+m)(k-m) = r (x - c)$
We next assume that $r = k - m$.
How do you then find $x$?

[nafistiham]

then you can say
$k+m=x-c$
or,$x=k+m+c$
or,$x=k-m+2m+c$
or,$x=r+2m+c$

[willpower]

Thanks a bunch! I finally understand.