Hard number theory

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td12345

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td12345

#1 h Apr 9, 2025, 11:32 PM

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Let $p$ be a prime number. Find the number of elements of the set
$M_p = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2p^{n} x} \in \mathbb{Q} \right\}.$
Edit: I edited with a variation of the same problem because I intend to give this one to an exam for my pupils.

This post has been edited 1 time. Last edited by td12345, Apr 14, 2025, 6:17 PM
Reason: variation of the problem

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mathprodigy2011

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#2 h Apr 10, 2025, 12:34 AM

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Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

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td12345

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td12345

#3 h Apr 10, 2025, 1:47 AM

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Regarding the counting of $M_2$ , didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$ . Can you please double check if you counted these in your solution ?

@math

mathprodigy2011 wrote:

Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 1:57 AM
Reason: typo

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mathprodigy2011

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mathprodigy2011

#4 h Apr 10, 2025, 2:51 AM

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td12345 wrote:

Regarding the counting of $M_2$ , didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$ . Can you please double check if you counted these in your solution ?

@math

mathprodigy2011 wrote:

Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned

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td12345

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td12345

#5 h Apr 10, 2025, 5:27 PM

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mathprodigy2011 wrote:

td12345 wrote:

Regarding the counting of $M_2$ , didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$ . Can you please double check if you counted these in your solution ?

@math

mathprodigy2011 wrote:

Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned

Yes, they lead to the same one when computing the square root, but the set counts the $x$ not the set of the square root expression no? I got a different number from yours and I don't know where I went wrong, I will lay down the $x$ I found as it goes over your claimed result and maybe one of us can figure out:

For $M_2$ :
- $x$ can be $(2^{2025-k}-2^{k-1})^2$ where $k \in \{1,2,\dots, 2025\}$ . This leads to $[\frac{2025+1}{2}]$ values of $x$ .
- $x$ can be $-(2^{2 \cdot 2025-2k}+2^{2k-2}+2^{2025})$ where $k \in \{1,2,\dots, 2025\}$ . This leads to $[\frac{2025+1}{2}]$ values of $x$ .
- $x$ can be $2^{2 \cdot 2025-3}+2-2^{2025}, -2^{2 \cdot 2025 - 3} -2- 2^{2025}$ . This leads to $2$ extra values for $x$ distinct from the above.

For $M_{2027}$ :
- $x$ can be $\frac{2027^k+2027^{2 \cdot 2025 - k}}{2} -2027^{2025}$ where $k \in \{0, 1,2,\dots, 2 \cdot 2025\}$ . This leads to $2025$ distinct non-zero values of $x$ .
- $x$ can be $\frac{-2027^k-2027^{2 \cdot 2025 - k}}{2} -2027^{2025}$ where $k \in \{0, 1,2,\dots, 2 \cdot 2025\}$ . This leads to $2026$ distinct non-zero values of $x$ .
I double checked with W|A to see if they lead to perfect squares under the square root and they do, so unless I double counted, the answer should be $2026+2+2025 + 2026$ ?

Edit: ok there is a miscount for $M_2$ because when $k=1013$ it leads to $0$ for the very first family and $x$ can't be $0$ . So for $M_2$ we shall have $1012+1013+2=2027$ . But for $M_{2027}$ count seems solid? I got $2024+2+2025+2026$

This post has been edited 8 times. Last edited by td12345, Apr 10, 2025, 8:34 PM
Reason: tpyo

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#6 h Apr 10, 2025, 8:05 PM

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Ok the 2 solutions I mentioned above as being separate (for $M_2$ ) actually come from another family:

- $x$ can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$ , where $k$ goes from 1 to 2024. This leads to $1012$ extra values for $x$ distinct from the above.
- $x$ can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$ , where $k$ goes from 1 to 2024. This leads to $1012$ extra values for $x$ distinct from the above.

So in total for $M_2$ we should have $1012 +1013+1012+1012=4049$ . Which would make the total $8100$ . Can somebody else confirm this answer?

This post has been edited 2 times. Last edited by td12345, Apr 10, 2025, 8:40 PM
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mathprodigy2011

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#7 h Apr 10, 2025, 9:10 PM

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td12345 wrote:

Ok the 2 solutions I mentioned above as being separate (for $M_2$ ) actually come from another family:

- $x$ can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$ , where $k$ goes from 1 to 2024. This leads to $1012$ extra values for $x$ distinct from the above.
- $x$ can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$ , where $k$ goes from 1 to 2024. This leads to $1012$ extra values for $x$ distinct from the above.

So in total for $M_2$ we should have $1012 +1013+1012+1012=4049$ . Which would make the total $8100$ . Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong

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td12345

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#8 h Apr 10, 2025, 9:29 PM

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mathprodigy2011 wrote:

td12345 wrote:

Ok the 2 solutions I mentioned above as being separate (for $M_2$ ) actually come from another family:

- $x$ can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$ , where $k$ goes from 1 to 2024. This leads to $1012$ extra values for $x$ distinct from the above.
- $x$ can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$ , where $k$ goes from 1 to 2024. This leads to $1012$ extra values for $x$ distinct from the above.

So in total for $M_2$ we should have $1012 +1013+1012+1012=4049$ . Which would make the total $8100$ . Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong

Yes it's integers that are not 0. $\mathbb{N}$ is natural numbers (positive integers) and $\mathbb{N}^*$ is natural numbers without $0$ . Would you like to give it another try just to see if it's really $8100$ ?

This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 9:30 PM
Reason: question

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