2021 JMPSC Invitationals Problems/Problem 11

Problem

For some $n$ , the arithmetic progression $4,9,14,\ldots,n$ has exactly $36$ perfect squares. Find the maximum possible value of $n.$

Solution

First note that the integers in the given arithmetic progression are precisely the integers which leave a remainder of $4$ when divided by $5$ .

Suppose a perfect square $m^2$ is in this arithmetic progression. Observe that the remainders when $0^2$ , $1^2$ , $2^2$ , $3^2$ , and $4^2$ are divided by $5$ are $0$ , $1$ , $4$ , $4$ , and $1$ , respectively. Furthermore, for any integer $m$ , $(m+5)^2 = m^2 + 10m + 25 = m^2 + 5(2m + 5),$ and so $(m+5)^2$ and $m^2$ leave the same remainder when divided by $5$ . It follows that the perfect squares in this arithmetic progression are exactly the numbers of the form $(5k+2)^2$ and $(5k+3)^2$ , respectively.

Finally, the sequence of such squares is $(5\cdot 0 + 2)^2, (5\cdot 0 + 3)^2, (5\cdot 1 + 2)^2, (5\cdot 1 + 3)^2,\cdots.$

In particular, the first and second such squares are associated with $k=1$ , the third and fourth are associated with $k=2$ , and so on. It follows that the $37^{\text{th}}$ such number, which is associated with $k=18$ , is $(5\cdot 18 + 2)^2 = 92^2 = 9409.$

Therefore the arithmetic progression must not reach $8464$ . This means the desired answer is $\boxed{8459}.$ ~djmathman

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2021 JMPSC Invitationals Problems/Problem 11

Problem

Solution

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