Henry_2001 wrote: For positive integer $k,$ we say that it is a Taurus integer if we can delete one element from the set $M_k=\{1,2,\cdots,k\},$ such that the sum of remaining $k-1$ elements is a positive perfect square. For example, $7$ is a Taurus integer, because if we delete $3$ from $M_7=\{1,2,3,4,5,6,7\},$ the sum of remaining $6$ elements is $25,$ which is a positive perfect square. $(1)$ Determine whether $2021$ is a Taurus integer. $(2)$ For positive integer $n,$ determine the number of Taurus integers in $\{1,2,\cdots,n\}.$ a)Yes,$M_k-{1190}$ is a perfect square

Sprites wrote: Henry_2001 wrote: For positive integer $k,$ we say that it is a Taurus integer if we can delete one element from the set $M_k=\{1,2,\cdots,k\},$ such that the sum of remaining $k-1$ elements is a positive perfect square. For example, $7$ is a Taurus integer, because if we delete $3$ from $M_7=\{1,2,3,4,5,6,7\},$ the sum of remaining $6$ elements is $25,$ which is a positive perfect square. $(1)$ Determine whether $2021$ is a Taurus integer. $(2)$ For positive integer $n,$ determine the number of Taurus integers in $\{1,2,\cdots,n\}.$ a)Yes,$M_k-{1190}$ is a perfect square b)We Claim that every integer is a tarus integer. Firstly,the sum of the set $M_n=\frac{n(n+1)}{2}>\lfloor \frac{n}{\sqrt{2}}\rfloor^2$ Now $\frac{n(n+1)}{2}-\frac{n^2}{2}<n $ hence every number is tarus . Sorry, but your answer of b) is wrong.(e.g. 5 is not Taurus)

tobylong wrote: Sprites wrote: Henry_2001 wrote: For positive integer $k,$ we say that it is a Taurus integer if we can delete one element from the set $M_k=\{1,2,\cdots,k\},$ such that the sum of remaining $k-1$ elements is a positive perfect square. For example, $7$ is a Taurus integer, because if we delete $3$ from $M_7=\{1,2,3,4,5,6,7\},$ the sum of remaining $6$ elements is $25,$ which is a positive perfect square. $(1)$ Determine whether $2021$ is a Taurus integer. $(2)$ For positive integer $n,$ determine the number of Taurus integers in $\{1,2,\cdots,n\}.$ a)Yes,$M_k-{1190}$ is a perfect square b)We Claim that every integer is a tarus integer. Firstly,the sum of the set $M_n=\frac{n(n+1)}{2}>\lfloor \frac{n}{\sqrt{2}}\rfloor^2$ Now $\frac{n(n+1)}{2}-\frac{n^2}{2}<n $ hence every number is tarus . Sorry, but your answer of b) is wrong.(e.g. 5 is not Taurus) Yeah,I guess it should work for all integers except $5$ and $8$

The answer is [sqrt((n^2+n-2)/2)]+1. Let f(n)=[sqrt((n^2+n-2)/2)].-[sqrt((n^2-n-2)/2)], it is easy to see that n is Taurus if and only if f(n)=1, and the answer is just to sum f(1)+f(2)+...+f(n). I think this is too easy for a 4.

Justanaccount wrote: The answer is $ \sqrt{\frac{n^2+n-2}{2}}+1$. Let $f(n)=\sqrt{\frac{n^2+n-2}{2}}- \sqrt{\frac{n^2-n-2}{2}}$, it is easy to see that n is Taurus if and only if f(n)=1, and the answer is just to sum $f(1)+f(2)+...+f(n)$. I think this is too easy for a 4. Can you post your sol please?