NT marathon!!!!!!!

For discussing Olympiad Level Number Theory problems
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Anindya Biswas
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Problem 5

Unread post by Anindya Biswas » Tue Mar 23, 2021 8:40 pm

Does there exists $2021$ positive integers whose sum of squares is also a perfect square?

Source :
Final exam of অনলাইন জীনতত্ত্ব ক্যাম্প 2015. :lol: (Final exam, ONT 2015, you will find the problemset in the National Math Camp thread).
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mehrab4226
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Re: Problem 5

Unread post by Mehrab4226 » Tue Mar 23, 2021 9:51 pm

Anindya Biswas wrote:
Tue Mar 23, 2021 8:40 pm
Does there exists $2021$ positive integers whose sum of squares is also a perfect square?

Source :
Final exam of অনলাইন জীনতত্ত্ব ক্যাম্প 2015. :lol: (Final exam, ONT 2015, you will find the problemset in the National Math Camp thread).
Do the 2021 integers have to be distinct?(Maybe they are distinct)
If not,
$(1^2+1^2+1^2+\cdots \text{2019 terms} \cdots +1^2+1^2)+4^2+9^2=46^2$
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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Anindya Biswas
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Re: Problem 5

Unread post by Anindya Biswas » Tue Mar 23, 2021 11:59 pm

Mehrab4226 wrote:
Tue Mar 23, 2021 9:51 pm
Anindya Biswas wrote:
Tue Mar 23, 2021 8:40 pm
Does there exists $2021$ positive integers whose sum of squares is also a perfect square?

Source :
Final exam of অনলাইন জীনতত্ত্ব ক্যাম্প 2015. :lol: (Final exam, ONT 2015, you will find the problemset in the National Math Camp thread).
Do the 2021 integers have to be distinct?(Maybe they are distinct)
If not,
$(1^2+1^2+1^2+\cdots \text{2019 terms} \cdots +1^2+1^2)+4^2+9^2=46^2$
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.
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

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Mehrab4226
Posts:230
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Location:Dhaka, Bangladesh

Solution of problem 5.1

Unread post by Mehrab4226 » Wed Mar 24, 2021 9:05 am

Anindya Biswas wrote:
Tue Mar 23, 2021 11:59 pm
Mehrab4226 wrote:
Tue Mar 23, 2021 9:51 pm
Anindya Biswas wrote:
Tue Mar 23, 2021 8:40 pm
Does there exists $2021$ positive integers whose sum of squares is also a perfect square?

Source :
Final exam of অনলাইন জীনতত্ত্ব ক্যাম্প 2015. :lol: (Final exam, ONT 2015, you will find the problemset in the National Math Camp thread).
Do the 2021 integers have to be distinct?(Maybe they are distinct)
If not,
$(1^2+1^2+1^2+\cdots \text{2019 terms} \cdots +1^2+1^2)+4^2+9^2=46^2$
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.
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".
We will solve this using Induction.
$\textbf{Base case:}$
$3^2=3^2$
$3^2+4^2=5^2$

$\textbf{Inductive Hypothesis:}$
Let, we can find $k$ distinct integers that add up to be a square of an odd integer. Just like the base case.

$\textbf{Inductive Step:}$
Let, $g^2$ be the sum of the squares of $k$ integers. By inductive Hypothesis $g$ is odd. So the square of $g$ is also odd. But we can write all odd numbers as the difference between $2$ consecutive square numbers. So let,

$g^2=p^2-q^2$
Here, $g$ is odd. And $p$ must also be odd. Because if it is not then in mod $4$ we get,
$1 \equiv -1 (mod4)$ [Which is definitely not possible.]
And again,
$p>g$ and,
$q=p-1$
And since g is odd,
$q>g$

So $q$ is not one of the $k$ square integers that make up $g^2$. So we got an odd square number which can be written a the sum of squares of $k+1$ distinct integers. $\square$
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

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Mehrab4226
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Problem: 6

Unread post by Mehrab4226 » 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} $
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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Anindya Biswas
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Re: Solution of problem 5.1

Unread post by Anindya Biswas » Wed Mar 24, 2021 9:33 am

Mehrab4226 wrote:
Wed Mar 24, 2021 9:05 am
Anindya Biswas wrote:
Tue Mar 23, 2021 11:59 pm
Mehrab4226 wrote:
Tue Mar 23, 2021 9:51 pm

Do the 2021 integers have to be distinct?(Maybe they are distinct)
If not,
$(1^2+1^2+1^2+\cdots \text{2019 terms} \cdots +1^2+1^2)+4^2+9^2=46^2$
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.
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".
We will solve is using Induction.
$\textbf{Base case:}$
$3^2=3^2$
$3^2+4^2=5^2$

$\textbf{Inductive Hypothesis:}$
Let, we can find $k$ distinct integers that add up to be a square of an odd integer. Just like the base case.

$\textbf{Inductive Step:}$
Let, $g^2$ be the sum of the squares of $k$ integers. By inductive Hypothesis $g$ is odd. So the square of $g$ is also odd. But we can write all odd numbers as the difference between $2$ consecutive square numbers. So let,

$g^2=p^2-q^2$
Here, $g$ is odd. And $p$ must also be odd. Because if it is not then in mod $4$ we get,
$1 \equiv -1 (mod4)$ [Which is definitely not possible.]
And again,
$p>g$ and,
$q=p-1$
And since g is odd,
$q>g$

So $q$ is not one of the $k$ square integers that make up $g$. So we got an odd square number which can be written a the sum of squares of $k+1$ distinct integers. $\square$
Wow! Brilliant...
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

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Anindya Biswas
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Re: Problem: 6

Unread post by Anindya Biswas » Wed Mar 24, 2021 11:47 am

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} $
For $p=2$,
$p^2+8=12$ which isn't a prime number.

For $p=3$,
$p^2+8=17$ and
$p^3+8p+2=53$ which are prime numbers.
So, the statement is true for $p=3$.

For $p>3$,
$p^2\equiv 1\pmod{3}$
$\Longrightarrow p^2+8\equiv0\pmod{3}$
Which means $p^2+8$ is not a prime number.

So we are done :D.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

abid.uhscian
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Re: NT marathon!!!!!!! problem-1

Unread post by abid.uhscian » Wed Mar 24, 2021 2:59 pm

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🙂

abid.uhscian
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Joined:Tue Mar 23, 2021 10:23 pm

Re: NT marathon!!!!!!! problem-1

Unread post by abid.uhscian » Wed Mar 24, 2021 3:07 pm

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🙂

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Anindya Biswas
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Problem 7

Unread post by Anindya Biswas » Wed Mar 24, 2021 3:26 pm

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 $12$
Source :
Slovenia 1995
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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