Mixing
Find the next least number N that is N+1 = X^2 and (N/2)+1 = y^2?
$begingroup$
To go with this following task to create a proper equation ?
Many numbers, especially from smaller ones, can rightly claim that
they are of particular interest to scientists. In the kingdom of the
right squares, the number 48 is particularly curious. If you add to it
1, you get a square [48 + 1 = 49 = 7x7], and if you split it half and
add 1 to the result, you will get another exact square [(48/2 = 24) +
1 = 25 = 5x5 ]. Both conditions are quite trivial, but together, they
are far less satisfied.
In fact, 48 is the smallest number that satisfies both conditions. Can
you find the next least number that meets both conditions?
linear-algebra sums-of-squares
asked Jan 10 at 9:23
AdelinAdelin
1013
$endgroup$
add a comment |
$begingroup$
To go with this following task to create a proper equation ?
Many numbers, especially from smaller ones, can rightly claim that
they are of particular interest to scientists. In the kingdom of the
right squares, the number 48 is particularly curious. If you add to it
1, you get a square [48 + 1 = 49 = 7x7], and if you split it half and
add 1 to the result, you will get another exact square [(48/2 = 24) +
1 = 25 = 5x5 ]. Both conditions are quite trivial, but together, they
are far less satisfied.
In fact, 48 is the smallest number that satisfies both conditions. Can
you find the next least number that meets both conditions?
linear-algebra sums-of-squares
asked Jan 10 at 9:23
AdelinAdelin
1013
$endgroup$
1
$begingroup$
The numbers can easier be found by searching for positive integers $n$ such that both $n$ and $2n-1$ are perfect squares. This boils down to find the positive integers $k$ such that $2k^2-1$ is a perfect square. This leads to a variant of the Pell-equation : $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 9:36
2
$begingroup$
The first $4$ numbers satisfying the conditions are : $$48 1680 57120 1940448$$
$endgroup$
– Peter
Jan 10 at 9:41
$begingroup$
@Peter why it is -1 not 0 ?
$endgroup$
– Adelin
Jan 10 at 13:25
$begingroup$
The square $m^2$ must be of the form $2k^2-1$, leading to $$m^2=2k^2-1$$ which is equivalent to $$m^2-2k^2=-1$$ or in other letters $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 13:27
1
$begingroup$
Also note that $(n/2+1)cdot 2-1=n+1$
$endgroup$
– Peter
Jan 10 at 13:28
add a comment |
$begingroup$
To go with this following task to create a proper equation ?
Many numbers, especially from smaller ones, can rightly claim that
they are of particular interest to scientists. In the kingdom of the
right squares, the number 48 is particularly curious. If you add to it
1, you get a square [48 + 1 = 49 = 7x7], and if you split it half and
add 1 to the result, you will get another exact square [(48/2 = 24) +
1 = 25 = 5x5 ]. Both conditions are quite trivial, but together, they
are far less satisfied.
In fact, 48 is the smallest number that satisfies both conditions. Can
you find the next least number that meets both conditions?
linear-algebra sums-of-squares
asked Jan 10 at 9:23
AdelinAdelin
1013
$endgroup$
To go with this following task to create a proper equation ?
Many numbers, especially from smaller ones, can rightly claim that
they are of particular interest to scientists. In the kingdom of the
right squares, the number 48 is particularly curious. If you add to it
1, you get a square [48 + 1 = 49 = 7x7], and if you split it half and
add 1 to the result, you will get another exact square [(48/2 = 24) +
1 = 25 = 5x5 ]. Both conditions are quite trivial, but together, they
are far less satisfied.
In fact, 48 is the smallest number that satisfies both conditions. Can
you find the next least number that meets both conditions?
linear-algebra sums-of-squares
linear-algebra sums-of-squares
asked Jan 10 at 9:23
AdelinAdelin
1013
asked Jan 10 at 9:23
AdelinAdelin
1013
edited Jan 10 at 9:42
Adelin
asked Jan 10 at 9:23
AdelinAdelin
1013
asked Jan 10 at 9:23
AdelinAdelin
1013
asked Jan 10 at 9:23
AdelinAdelin
1013
1013
1
$begingroup$
The numbers can easier be found by searching for positive integers $n$ such that both $n$ and $2n-1$ are perfect squares. This boils down to find the positive integers $k$ such that $2k^2-1$ is a perfect square. This leads to a variant of the Pell-equation : $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 9:36
2
$begingroup$
The first $4$ numbers satisfying the conditions are : $$48 1680 57120 1940448$$
$endgroup$
– Peter
Jan 10 at 9:41
$begingroup$
@Peter why it is -1 not 0 ?
$endgroup$
– Adelin
Jan 10 at 13:25
$begingroup$
The square $m^2$ must be of the form $2k^2-1$, leading to $$m^2=2k^2-1$$ which is equivalent to $$m^2-2k^2=-1$$ or in other letters $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 13:27
1
$begingroup$
Also note that $(n/2+1)cdot 2-1=n+1$
$endgroup$
– Peter
Jan 10 at 13:28
add a comment |
1
$begingroup$
The numbers can easier be found by searching for positive integers $n$ such that both $n$ and $2n-1$ are perfect squares. This boils down to find the positive integers $k$ such that $2k^2-1$ is a perfect square. This leads to a variant of the Pell-equation : $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 9:36
2
$begingroup$
The first $4$ numbers satisfying the conditions are : $$48 1680 57120 1940448$$
$endgroup$
– Peter
Jan 10 at 9:41
$begingroup$
@Peter why it is -1 not 0 ?
$endgroup$
– Adelin
Jan 10 at 13:25
$begingroup$
The square $m^2$ must be of the form $2k^2-1$, leading to $$m^2=2k^2-1$$ which is equivalent to $$m^2-2k^2=-1$$ or in other letters $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 13:27
1
$begingroup$
Also note that $(n/2+1)cdot 2-1=n+1$
$endgroup$
– Peter
Jan 10 at 13:28
1
1
$begingroup$
The numbers can easier be found by searching for positive integers $n$ such that both $n$ and $2n-1$ are perfect squares. This boils down to find the positive integers $k$ such that $2k^2-1$ is a perfect square. This leads to a variant of the Pell-equation : $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 9:36
$begingroup$
The numbers can easier be found by searching for positive integers $n$ such that both $n$ and $2n-1$ are perfect squares. This boils down to find the positive integers $k$ such that $2k^2-1$ is a perfect square. This leads to a variant of the Pell-equation : $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 9:36
2
2
$begingroup$
The first $4$ numbers satisfying the conditions are : $$48 1680 57120 1940448$$
$endgroup$
– Peter
Jan 10 at 9:41
$begingroup$
The first $4$ numbers satisfying the conditions are : $$48 1680 57120 1940448$$
$endgroup$
– Peter
Jan 10 at 9:41
$begingroup$
@Peter why it is -1 not 0 ?
$endgroup$
– Adelin
Jan 10 at 13:25
$begingroup$
@Peter why it is -1 not 0 ?
$endgroup$
– Adelin
Jan 10 at 13:25
$begingroup$
The square $m^2$ must be of the form $2k^2-1$, leading to $$m^2=2k^2-1$$ which is equivalent to $$m^2-2k^2=-1$$ or in other letters $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 13:27
$begingroup$
The square $m^2$ must be of the form $2k^2-1$, leading to $$m^2=2k^2-1$$ which is equivalent to $$m^2-2k^2=-1$$ or in other letters $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 13:27
1
1
$begingroup$
Also note that $(n/2+1)cdot 2-1=n+1$
$endgroup$
– Peter
Jan 10 at 13:28
$begingroup$
Also note that $(n/2+1)cdot 2-1=n+1$
$endgroup$
– Peter
Jan 10 at 13:28
add a comment |
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1
$begingroup$
The numbers can easier be found by searching for positive integers $n$ such that both $n$ and $2n-1$ are perfect squares. This boils down to find the positive integers $k$ such that $2k^2-1$ is a perfect square. This leads to a variant of the Pell-equation : $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 9:36
2
$begingroup$
The first $4$ numbers satisfying the conditions are : $$48 1680 57120 1940448$$
$endgroup$
– Peter
Jan 10 at 9:41
$begingroup$
@Peter why it is -1 not 0 ?
$endgroup$
– Adelin
Jan 10 at 13:25
$begingroup$
The square $m^2$ must be of the form $2k^2-1$, leading to $$m^2=2k^2-1$$ which is equivalent to $$m^2-2k^2=-1$$ or in other letters $$a^2-2b^2=-1$$
$endgroup$
– Peter
Jan 10 at 13:27
1
$begingroup$
Also note that $(n/2+1)cdot 2-1=n+1$
$endgroup$
– Peter
Jan 10 at 13:28