# 2021 JMPSC Invitationals Problems/Problem 11

## Problem

For some , the arithmetic progression has exactly perfect squares. Find the maximum possible value of

## Solution

First note that the integers in the given arithmetic progression are precisely the integers which leave a remainder of when divided by .

Suppose a perfect square is in this arithmetic progression. Observe that the remainders when , , , , and are divided by are , , , , and , respectively. Furthermore, for any integer , and so and leave the same remainder when divided by . It follows that the perfect squares in this arithmetic progression are exactly the numbers of the form and , respectively.

Finally, the sequence of such squares is

In particular, the first and second such squares are associated with , the third and fourth are associated with , and so on. It follows that the such number, which is associated with , is

Therefore the arithmetic progression must not reach . This means the desired answer is ~djmathman

## Solution 2

We examine all perfect squares ending in or are part of our sequence, so for every cycle of perfect squares, exactly are included. This means cycles are included, which goes until . Now, note is not part of our sequence, but is the th perfect square. Therefore, below this yields , which is the answer.

~Geometry285

## Solution 3

Since this arithmetic progression has common ratio 5. Thus, all terms in it are in the form Taking the modulo 5, all have either or Thus all integers of the form are in the arithmetic progression and are perfect squares. This means that the 37 perfect square in the progression is This also implies that the maximum value of is

~pinkpig

## See also

- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.