# I am not able to solve this question on set theory I tried but I am unsure of my answer

Let $$S$$ be the set of all positive integers such that each element $$n$$ in $$S$$ satisfies the following conditions:

1. The sum of the digits of $$n$$ is a prime number.
2. The sum of the digits of $$n^2$$ is a perfect square.

Prove or disprove the existence of an infinite subset $$T$$ of $$S$$ such that for every $$n$$ in $$T$$, the sum of the digits of $$n^3$$ is a perfect cube.

Attempt

To prove the existence of an infinite subset $$T$$ of $$S$$ such that for every $$n$$ in $$T$$, the sum of the digits of $$n^3$$ is a perfect cube, we need to show that there are infinitely many positive integers $$n$$ satisfying the given conditions.

I analyzed the conditions provided:

1. The sum of the digits of $$n$$ is a prime number.
2. The sum of the digits of $$n^2$$ is a perfect square.

I tried to construct such an infinite subset $$T$$ by considering the properties of the conditions.

First, i started with the condition that the sum of the digits of $$n$$ is a prime number. This condition restricts the possible values of $$n$$ to those whose digit sums are primes. Since there are infinitely many prime numbers, there are infinitely many such $$n$$.

Second, i considered the condition that the sum of the digits of $$n^2$$ is a perfect square. For any given prime digit sum $$p$$, there are infinitely many numbers whose digit sum is $$p$$. Let's denote one such number as $$n_1$$. Now, if $$n_1$$ satisfies the condition that the sum of the digits of $$n_1^2$$ is a perfect square, we can include it in our subset $$T$$.

Now if I continue this process to construct infinitely many numbers in $$T$$, each satisfying the given conditions. Since there are infinitely many prime numbers and infinitely many numbers for each prime digit sum, we can conclude that there exists an infinite subset$$T$$ of $$S$$ satisfying the condition that the sum of the digits of $$n^3$$ is a perfect cube for every $$n$$ in $$T$$.

Therefore, can we say that the existence of such infinite subset $$T$$ is proven?

I am unsure of my answer. Is this correct?

• Note: I don't understand your argument. You appear to simply assert that you can find natural numbers with some desired properties, but this isn't obvious.
– lulu
Commented Mar 18 at 14:22
• The set $T = \{2 \times 10^n\}$ indeed gets the job done, as $2$ is a prime, $4$ is a perfect square, and $8$ is a perfect cube.
– D S
Commented Mar 18 at 14:27
• @DS Not sure what I was thinking there. Thanks! Of course $\{2\times 10^n\}$ works, as you point out.
– lulu
Commented Mar 18 at 14:28
• To the OP: Your attempt is flawed. If $A$ is infinite, $B$ is infinite, then $A \cap B$ is not necessarily infinite.
– D S
Commented Mar 18 at 14:29
• @DS It's easier than that because $11$ works as well. In fact, I believe you can construct numbers with two ones separated by zeroes so that the square has non-zero digits $1,2,1$ and the digits of the cube are $1,3,3,1$. That would give the infinite set. Commented Mar 18 at 14:52

Numbers of the form $$10^n+1$$ work for all positive $$n$$.

The digits of $$10^n+1$$ sum to $$2$$.

$$(10^n+1)^2=10^{2n}+2\times 10^n+1$$ so the digits sum to $$4$$.

$$(10^n+1)^3=10^{3n}+3\times 10^{2n}+3\times 10^n+1$$ and the digits sum to $$8$$.

• Also, $(10^n+1)10^k$ work too.
– D S
Commented Mar 18 at 17:59
• I still don't get the point you were trying to make in the comments. Doesn't the set $\{2,20,200,2000\ldots \}$ work too?
– D S
Commented Mar 18 at 18:00
• @DS Yes, it does which shows that there are more than one. Commented Mar 18 at 18:55
• No, $p=5$doesn't work, neither does p = 7.
– D S
Commented Mar 18 at 21:20