N is a good number, if in any subset of M of size n there are two elements a and b such that a+b is a prime number

[lfdahl]

$M$ is the set of squares of the fi rst $20$ natural numbers:\[M = \left\{1^2,2^2,3^2,...,19^2,20^2\right\}\]We say that $n$ is a good number, if in any subset of $M$ of size $n$ there are two
elements $a$ and $b$ such that $a + b$ is a prime number. Find the smallest good number.

 

[castor28]

$M$ is the set of squares of the first $20$ natural numbers:\[M = \left\{1^2,2^2,3^2,...,19^2,20^2\right\}\]We say that $n$ is a good number, if in any subset of $M$ of size $n$ there are two
elements $a$ and $b$ such that $a + b$ is a prime number. Find the smallest good number.

I'm not quite sure about this
[sp]

We observe first that, if we have a subset $S\subset M$ of $k$ elements such that all the sums of pairs of elements of $S$ are composite (we will call that a bad subset), then $n > k$, where $n$ is the smallest good number. Taking $S$ as the set of $10$ even squares, we see that $n\ge11$.

To prove that $n=11$, we must show that there is no bad subset of $11$ elements. As the sum of any two distinct elements of the same parity is composite, we need only consider the sums of an odd and an even number.

The following table contains all the possible such sums of squares.
$$
\begin{array}{r|rrrrrrrrrr|r}
& 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 & 17 & 19 & t_r\\
\hline
2 & 5 & 13 & 29 & 53 & \bf 85 & \bf 125 & 173 & 229 & 293 & \bf 365 & 3 \\
4 & 17 & \bf 25 & 41 & \bf 65 & 97 & 137 & \bf 185 & 241 & \bf 305 & \bf 377 & 5 \\
6 & 37 & \bf 45 & 61 & \bf 85 & \bf 117 & 157 & \bf 205 & \bf 261 & \bf 325 & 397 & 6 \\
8 & \bf 65 & 73 & 89 & 113 & \bf 145 & \bf 185 & 233 & \bf 289 & 353 & \bf 425 & 5 \\
10 & 101 & 109 & \bf 125 & 149 & 181 & \bf 221 & 269 & \bf 325 & 389 & 461 & 3 \\
12 & \bf 145 & \bf 153 & \bf 169 & 193 & \bf 225 & \bf 265 & 313 & \bf 369 & 433 & \bf 505 & 7 \\
14 & 197 & \bf 205 & \bf 221 & \bf 245 & 277 & 317 & \bf 365 & 421 & \bf 485 & 557 & 5 \\
16 & 257 & \bf 265 & 281 & \bf 305 & 337 & \bf 377 & \bf 425 & \bf 481 & \bf 545 & 617 & 6 \\
20 & 401 & 409 & \bf 425 & 449 & \bf 481 & 521 & 569 & \bf 625 & \bf 689 & 761 & 4 \\
\hline
t_c & 2 & 5 & 4 & 4 & 5 & 5 & 4 & 6 & 5 & 4
\end{array}
$$
The composite numbers are in bold; the last row ($t_c$) and the last column ($t_r$) contain the numbers of composite numbers in the corresponding columns and rows, respectively.

Assume that there exists a bad subset of $11$ elements; let $n_c$ and $n_r$ be the numbers of rows (even numbers) and columns (odd numbers) in that subset; $n_c + n_r = 11$. We must select $n_r$ rows and $n_c$ columns in such a way that all the remaining entries are composite.

As the row with the largest $t_r$ contains $7$ composite numbers, we must have $n_c \le 7$; this implies that we must have at least $4$ rows.

As the fourth larges value of $t_r$ is $5$, we must have $n_c\le5$, and therefore $n_r\ge6$. As the sixth largest $t_r$ is $5$ again, this gives nothing new.

As the largest column total $t_c$ is $6$, we must have $n_r\le6$; this means that we must have $n_r=6$ and $n_c=5$. Now, as the fifth largest $t_c$ is $5$, we must have $n_r\le5$, and this is a contradiction.

We conclude that there can be no bad subset of $11$ elements, and $n=11$.

[/sp]

 

[lfdahl]

I'm not quite sure about this
[sp]

We observe first that, if we have a subset $S\subset M$ of $k$ elements such that all the sums of pairs of elements of $S$ are composite (we will call that a bad subset), then $n > k$, where $n$ is the smallest good number. Taking $S$ as the set of $10$ even squares, we see that $n\ge11$.

To prove that $n=11$, we must show that there is no bad subset of $11$ elements. As the sum of any two distinct elements of the same parity is composite, we need only consider the sums of an odd and an even number.

The following table contains all the possible such sums of squares.
$$
\begin{array}{r|rrrrrrrrrr|r}
& 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 & 17 & 19 & t_r\\
\hline
2 & 5 & 13 & 29 & 53 & \bf 85 & \bf 125 & 173 & 229 & 293 & \bf 365 & 3 \\
4 & 17 & \bf 25 & 41 & \bf 65 & 97 & 137 & \bf 185 & 241 & \bf 305 & \bf 377 & 5 \\
6 & 37 & \bf 45 & 61 & \bf 85 & \bf 117 & 157 & \bf 205 & \bf 261 & \bf 325 & 397 & 6 \\
8 & \bf 65 & 73 & 89 & 113 & \bf 145 & \bf 185 & 233 & \bf 289 & 353 & \bf 425 & 5 \\
10 & 101 & 109 & \bf 125 & 149 & 181 & \bf 221 & 269 & \bf 325 & 389 & 461 & 3 \\
12 & \bf 145 & \bf 153 & \bf 169 & 193 & \bf 225 & \bf 265 & 313 & \bf 369 & 433 & \bf 505 & 7 \\
14 & 197 & \bf 205 & \bf 221 & \bf 245 & 277 & 317 & \bf 365 & 421 & \bf 485 & 557 & 5 \\
16 & 257 & \bf 265 & 281 & \bf 305 & 337 & \bf 377 & \bf 425 & \bf 481 & \bf 545 & 617 & 6 \\
20 & 401 & 409 & \bf 425 & 449 & \bf 481 & 521 & 569 & \bf 625 & \bf 689 & 761 & 4 \\
\hline
t_c & 2 & 5 & 4 & 4 & 5 & 5 & 4 & 6 & 5 & 4
\end{array}
$$
The composite numbers are in bold; the last row ($t_c$) and the last column ($t_r$) contain the numbers of composite numbers in the corresponding columns and rows, respectively.

Assume that there exists a bad subset of $11$ elements; let $n_c$ and $n_r$ be the numbers of rows (even numbers) and columns (odd numbers) in that subset; $n_c + n_r = 11$. We must select $n_r$ rows and $n_c$ columns in such a way that all the remaining entries are composite.

As the row with the largest $t_r$ contains $7$ composite numbers, we must have $n_c \le 7$; this implies that we must have at least $4$ rows.

As the fourth larges value of $t_r$ is $5$, we must have $n_c\le5$, and therefore $n_r\ge6$. As the sixth largest $t_r$ is $5$ again, this gives nothing new.

As the largest column total $t_c$ is $6$, we must have $n_r\le6$; this means that we must have $n_r=6$ and $n_c=5$. Now, as the fifth largest $t_c$ is $5$, we must have $n_r\le5$, and this is a contradiction.

We conclude that there can be no bad subset of $11$ elements, and $n=11$.

[/sp]

Thankyou, castor28, for your careful analysis and correct solution. Your approach works in my view. I´m glad, that you participated in this challenge. (Yes)

The suggested solution demonstrates a shortcut:

Spoiler

The answer is $n = 11$.
Let $K = \left \{ 1^2,3^2,5^2,...,19^2 \right \}$. Since the sum of any two elements of is not prime, $n \geq 11$.
Now let us show, that $n \leq 11$. We partition $M$ into $10$ subsets of order $2$ such that the sum of two elements in any subset is prime:

\[M = \left \{ 1^2,4^2 \right \}\cup \left \{2^2,3^2 \right \}\cup \left \{5^2,8^2 \right \}\cup \left \{ 6^2,11^2\right \}\cup \left \{7^2,10^2 \right \}\cup \left \{9^2,16^2 \right \}\cup \left \{12^2,13^2 \right \}\cup \left \{14^2,15^2 \right \}\cup \left \{17^2,18^2 \right \}\cup \left \{19^2,20^2 \right \}.\]

Any subset of $M$ having $11$ elements contains both elements of at least one of these $10$ subsets. Done.