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NT marathon!!!!!!!
- Anindya Biswas
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Does there exists $2021$ positive integers whose sum of squares is also a perfect square?
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"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
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Re: Problem 5
Do the 2021 integers have to be distinct?(Maybe they are distinct)Anindya Biswas wrote: ↑Tue Mar 23, 2021 8:40 pmDoes there exists $2021$ positive integers whose sum of squares is also a perfect square?
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If not,
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
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Re: Problem 5
The real question was a mess, they only asked for integers. I used "positive integers" term so that we can't use $0$. Try to show it for distinct integers.Mehrab4226 wrote: ↑Tue Mar 23, 2021 9:51 pmDo the 2021 integers have to be distinct?(Maybe they are distinct)Anindya Biswas wrote: ↑Tue Mar 23, 2021 8:40 pmDoes there exists $2021$ positive integers whose sum of squares is also a perfect square?
Source :
If not,
I have a more general statement, "For every $n\in\mathbb{N}$ there exists $n$ distinct positive integers whose sum of squares is also a perfect square".
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Mehrab4226
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Solution of problem 5.1
Anindya Biswas wrote: ↑Tue Mar 23, 2021 11:59 pmThe real question was a mess, they only asked for integers. I used "positive integers" term so that we can't use $0$. Try to show it for distinct integers.Mehrab4226 wrote: ↑Tue Mar 23, 2021 9:51 pmDo the 2021 integers have to be distinct?(Maybe they are distinct)Anindya Biswas wrote: ↑Tue Mar 23, 2021 8:40 pmDoes there exists $2021$ positive integers whose sum of squares is also a perfect square?
Source :
If not,
I have a more general statement, "For every $n\in\mathbb{N}$ there exists $n$ distinct positive integers whose sum of squares is also a perfect square".
Last edited by Mehrab4226 on Wed Mar 24, 2021 7:25 pm, edited 2 times in total.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
- Mehrab4226
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Problem: 6
$\text{Prove that, if }p\text{ and }p^2+8 \text{ are primes, then, }p^3+8p+2 \text{ is prime} $
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
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Re: Solution of problem 5.1
Wow! Brilliant...Mehrab4226 wrote: ↑Wed Mar 24, 2021 9:05 amAnindya Biswas wrote: ↑Tue Mar 23, 2021 11:59 pmThe real question was a mess, they only asked for integers. I used "positive integers" term so that we can't use $0$. Try to show it for distinct integers.Mehrab4226 wrote: ↑Tue Mar 23, 2021 9:51 pm
Do the 2021 integers have to be distinct?(Maybe they are distinct)
If not,
I have a more general statement, "For every $n\in\mathbb{N}$ there exists $n$ distinct positive integers whose sum of squares is also a perfect square".
I had another solution in my mind,
\[\left(a^2+2\left(b_1^2+b_2^2+\dots+b_n^2\right)\right)^2=\left(a^2\right)^2+\left(2b_1^2+2b_2^2+\dots+2b_n^2\right)^2+\left(2ab_1\right)^2+\left(2ab_2\right)^2+\dots+\left(2ab_n\right)^2\]
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Anindya Biswas
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Re: Problem: 6
Mehrab4226 wrote: ↑Wed Mar 24, 2021 9:22 am$\text{Prove that, if }p\text{ and }p^2+8 \text{ are primes, then, }p^3+8p+2 \text{ is prime} $
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
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Re: NT marathon!!!!!!! problem-1
X is odd and Y is an integer
Now,
X^3 Ξ 0,1,2 (mod 3)
And 2Y^2+1 Ξ 0,1 (mod 3)
In the equation if we put the value of X as 0 or 1 or 2 and Y as 0 or 1, we get only one pair of (X,Y)=(1,0) for which the LHS and the RHS is equal.
Thus the, X^3|2Y^2+1
But 0 isn't a positive integer. That's why there is no solution in the pair of positive integers
If I am wrong please share me the correct solution because I am a beginner
Now,
X^3 Ξ 0,1,2 (mod 3)
And 2Y^2+1 Ξ 0,1 (mod 3)
In the equation if we put the value of X as 0 or 1 or 2 and Y as 0 or 1, we get only one pair of (X,Y)=(1,0) for which the LHS and the RHS is equal.
Thus the, X^3|2Y^2+1
But 0 isn't a positive integer. That's why there is no solution in the pair of positive integers
If I am wrong please share me the correct solution because I am a beginner
-
- Posts:3
- Joined:Tue Mar 23, 2021 10:23 pm
Re: NT marathon!!!!!!! problem-1
X is odd and Y is an integer
Now,
X^3 Ξ 0,1,2 (mod 3)
And 2Y^2+1 Ξ 0,1 (mod 3)
In the equation if we put the value of X as 0 or 1 or 2 and Y as 0 or 1, we get only one pair of (X,Y)=(1,0) for which the LHS and the RHS is equal.
Thus the, X^3|2Y^2+1
But 0 isn't a positive integer. That's why there is no solution in positive integers
If I am wrong please share me the correct solution because I am a beginner
Now,
X^3 Ξ 0,1,2 (mod 3)
And 2Y^2+1 Ξ 0,1 (mod 3)
In the equation if we put the value of X as 0 or 1 or 2 and Y as 0 or 1, we get only one pair of (X,Y)=(1,0) for which the LHS and the RHS is equal.
Thus the, X^3|2Y^2+1
But 0 isn't a positive integer. That's why there is no solution in positive integers
If I am wrong please share me the correct solution because I am a beginner
- Anindya Biswas
- Posts:264
- Joined:Fri Oct 02, 2020 8:51 pm
- Location:Magura, Bangladesh
- Contact:
Problem 7
Let $a,b, c, d$ be integers. Show that the product \[(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)\] is divisible by $12$
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"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann