Solution

a) Note that 48 is a term in the sequence that is a multiple of 3 and satisfies $48+1=49=7^{2}$. Thus, to find its successor, we will try to obtain a multiple of three that, when increased by one unit, equals $8^{2}=64$. Subtracting one unit from 64, we get 63, which is indeed a multiple of 3.

In other words, we have $63+1=64=8^{2}$, so 63 is the next term in the sequence.

Note: Note that the same reasoning does not work to find the term in the sequence that appears after the term 63. Indeed, if we calculate $9^{2}-1=81-1=80$, we end up finding a number that is not a multiple of 3. The procedure developed in items b) and c) below shows us how to overcome this type of difficulty.

b) We start by writing the first perfect squares in ascending order:

$$
2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}, 8^{2}, 9^{2}, 10^{2}, 11^{2}, 12^{2}, 13^{2}, 14^{2}, 15^{2}, \ldots
$$

which is equal to the sequence

$$
4,9,16,25,36,49,64,81,100,121,144,169,196,225, \ldots
$$

Subtracting one unit from each of the terms, we get the new sequence

$$
3,8,15,24,35,48,63,80,99,120,143,168,195,224, \ldots
$$

We exclude from this list those numbers that are not multiples of 3, obtaining the list:

$$
3,15,24,48,63,99,120,168,195, \ldots
$$

This is then an increasing list of multiples of 3 whose sum with 1 results in a perfect square. Note that its eighth term is equal to 168. Therefore, this is the number we are looking for.

Note: We notice from the solution above that when we subtract one unit from the square of any number that is not a multiple of 3, we get a multiple of 3 as a result.

For example, the number 4 is not a multiple of 3, yet $4^{2}-1=16-1=15$ which is a multiple of 3. Also, 11 is not a multiple of 3, but $11^{2}-1=121-1=120$ which is a multiple of 3.

Note that the multiples of 3 are the numbers $x$ of the form $x=3n$ for some natural number $n$. The numbers that are not multiples of 3 are those that can be written in the form $x=3n+1$ or $x=3n+2$. These two types of numbers correspond respectively to those whose remainder when divided by 3 is equal to 1 or 2.

Consider first a number $x$ of the first type, i.e., of the form $x=3n+1$. Note that $x^{2}=(3n+1)^{2}=9n^{2}+6n+1$. Thus, $x^{2}-1=9n^{2}+6n$ which is a multiple of 3.

Similarly, for numbers of the second type $x=3n+2$, we have $x^{2}=9n^{2}+12n+4$, so $x^{2}-1=9n^{2}+12n+3$ which is also a multiple of 3.

Finally, note that for a multiple of $3, x=3n$, it holds that $x^{2}-1=9n^{2}-1$ which cannot be a multiple of 3, since $9n^{2}$ is a multiple of 3 by itself.

c) We are interested in calculating the $2013^{\text{th}}$ term of the list:

$$
3,15,24,48,63,99,120,168,195, \ldots
$$

which appears in the previous item. Let the numbers of type 1 and type 2 be those whose division by 3 leaves a remainder equal to 2 and 1, respectively. From the observation above, we conclude that to generate this list, we can first form an increasing list by alternating numbers of type 1 and type 2. Then we must square each term and, finally, subtract one unit from each term.

Since we are interested in finding the $2013^{\text{th}}$ number, we can focus on finding the $1007^{\text{th}}$ term of the sequence of numbers of type 1. The first term of this list is equal to $3 \times 0+2=2$, the second is equal to $3 \times 1+2=5$, the third is equal to $3 \times 2+2=8$, and so on. Therefore, the $1007^{\text{th}}$ term is equal to $3 \times 1006+2=3020$. Thus, to generate the $2013^{\text{th}}$ term of the list

$$
3,15,24,48,63,99,120,168,195, \ldots
$$

we need to square the number 3020 and subtract one unit, obtaining the number 9120399.
Write, in ascending order, each of the multiples of 3 whose sum with 1 is a perfect square:

$$
3,15,24,48, \ldots
$$

a) What is the next number that will appear in this sequence after 48?

b) What is the eighth number in this sequence?

c) What is the number that will appear in this sequence in the $2013^{th}$ position?