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".
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. $\square$
Problem: 6
Posted: Wed Mar 24, 2021 9:22 am
by Mehrab4226
$\text{Prove that, if }p\text{ and }p^2+8 \text{ are primes, then, }p^3+8p+2 \text{ is prime} $
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,
\[\left(a^2+2\left(b_1^2+b_2^2+\dots+b_n^2\right)\right)^2=\left(a^2\right)^2+\left(2b_1^2+2b_2^2+\dots+2b_n^2\right)^2+\left(2ab_1\right)^2+\left(2ab_2\right)^2+\dots+\left(2ab_n\right)^2\]
$\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 .
Re: NT marathon!!!!!!! problem-1
Posted: Wed Mar 24, 2021 2:59 pm
by abid.uhscian
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
Re: NT marathon!!!!!!! problem-1
Posted: Wed Mar 24, 2021 3:07 pm
by abid.uhscian
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
Problem 7
Posted: Wed Mar 24, 2021 3:26 pm
by Anindya Biswas
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 :
(Final exam, ONT 2015, you will find the problemset in the National Math Camp thread).
.