Mixing
Beautiful Triangles
$begingroup$
Let $p_1$,$p_2$,...,$p_{2017}$, be the first 2017 odd prime numbers and none are equal. Let n=product of all the primes . A triangle is a "beautiful triangle" if it is a right triangle, has sides' lengths are integers and the radius of the circumcircle of the triangle is n. Let X be the number of "beautiful triangles". Find the last 2 digits of X.
This is a tough problem, but i tried a little.
If it is a right angled triangle, then the perpendicular bisector should be parallel to the perpendicular side. Also, the circumcenter should lie somewhere on this line. Then if the radius is n, let us assume the base side can be broken into two-x and x(since perpendicular bisector). Then x can be written in terms of n. But the most difficult part is- what to do next.
Please help me!
combinatorics triangle
$endgroup$
|
show 5 more comments
$begingroup$
Let $p_1$,$p_2$,...,$p_{2017}$, be the first 2017 odd prime numbers and none are equal. Let n=product of all the primes . A triangle is a "beautiful triangle" if it is a right triangle, has sides' lengths are integers and the radius of the circumcircle of the triangle is n. Let X be the number of "beautiful triangles". Find the last 2 digits of X.
This is a tough problem, but i tried a little.
If it is a right angled triangle, then the perpendicular bisector should be parallel to the perpendicular side. Also, the circumcenter should lie somewhere on this line. Then if the radius is n, let us assume the base side can be broken into two-x and x(since perpendicular bisector). Then x can be written in terms of n. But the most difficult part is- what to do next.
Please help me!
combinatorics triangle
$endgroup$
1
$begingroup$
"Let p₁,p₂,...,p₂₀₁₇, be the 2017 odd prime numbers and none are equal." -- Which primes, because there are infinitely many primes, almost all odd except $2$? Similarly -- "Let n=product of all the primes" -- this product is infinite. Is this referring to the product $p_1 times ... times p_{2017}$?
$endgroup$
– Eevee Trainer
Jan 21 at 16:33
$begingroup$
Yes sir i could not type it
$endgroup$
– user636268
Jan 21 at 16:34
$begingroup$
What does it mean for the radius of the radius of the circumcircle to be in $n$? Does it divide $n$? For a right triangle the radius of the circumcircle is half the hypotenuse.
$endgroup$
– Ross Millikan
Jan 21 at 16:54
$begingroup$
Yes, n is the product of all the given primes
$endgroup$
– user636268
Jan 21 at 16:56
1
$begingroup$
So if we let $r= 2n/13,$ for example, we can show that the right triangle with legs $5r$ and $12r$ is beautiful. So I think it comes down to the number (modulo $100$) of primitive Pythagorean triples in which the hypotenuse is a product of some subset of the first $2017$ odd primes.
$endgroup$
– David K
Jan 21 at 17:26
|
show 5 more comments
$begingroup$
Let $p_1$,$p_2$,...,$p_{2017}$, be the first 2017 odd prime numbers and none are equal. Let n=product of all the primes . A triangle is a "beautiful triangle" if it is a right triangle, has sides' lengths are integers and the radius of the circumcircle of the triangle is n. Let X be the number of "beautiful triangles". Find the last 2 digits of X.
This is a tough problem, but i tried a little.
If it is a right angled triangle, then the perpendicular bisector should be parallel to the perpendicular side. Also, the circumcenter should lie somewhere on this line. Then if the radius is n, let us assume the base side can be broken into two-x and x(since perpendicular bisector). Then x can be written in terms of n. But the most difficult part is- what to do next.
Please help me!
combinatorics triangle
$endgroup$
Let $p_1$,$p_2$,...,$p_{2017}$, be the first 2017 odd prime numbers and none are equal. Let n=product of all the primes . A triangle is a "beautiful triangle" if it is a right triangle, has sides' lengths are integers and the radius of the circumcircle of the triangle is n. Let X be the number of "beautiful triangles". Find the last 2 digits of X.
This is a tough problem, but i tried a little.
If it is a right angled triangle, then the perpendicular bisector should be parallel to the perpendicular side. Also, the circumcenter should lie somewhere on this line. Then if the radius is n, let us assume the base side can be broken into two-x and x(since perpendicular bisector). Then x can be written in terms of n. But the most difficult part is- what to do next.
Please help me!
combinatorics triangle
combinatorics triangle
edited Jan 21 at 17:17
asked Jan 21 at 16:30
user636268
1
$begingroup$
"Let p₁,p₂,...,p₂₀₁₇, be the 2017 odd prime numbers and none are equal." -- Which primes, because there are infinitely many primes, almost all odd except $2$? Similarly -- "Let n=product of all the primes" -- this product is infinite. Is this referring to the product $p_1 times ... times p_{2017}$?
$endgroup$
– Eevee Trainer
Jan 21 at 16:33
$begingroup$
Yes sir i could not type it
$endgroup$
– user636268
Jan 21 at 16:34
$begingroup$
What does it mean for the radius of the radius of the circumcircle to be in $n$? Does it divide $n$? For a right triangle the radius of the circumcircle is half the hypotenuse.
$endgroup$
– Ross Millikan
Jan 21 at 16:54
$begingroup$
Yes, n is the product of all the given primes
$endgroup$
– user636268
Jan 21 at 16:56
1
$begingroup$
So if we let $r= 2n/13,$ for example, we can show that the right triangle with legs $5r$ and $12r$ is beautiful. So I think it comes down to the number (modulo $100$) of primitive Pythagorean triples in which the hypotenuse is a product of some subset of the first $2017$ odd primes.
$endgroup$
– David K
Jan 21 at 17:26
|
show 5 more comments
1
$begingroup$
"Let p₁,p₂,...,p₂₀₁₇, be the 2017 odd prime numbers and none are equal." -- Which primes, because there are infinitely many primes, almost all odd except $2$? Similarly -- "Let n=product of all the primes" -- this product is infinite. Is this referring to the product $p_1 times ... times p_{2017}$?
$endgroup$
– Eevee Trainer
Jan 21 at 16:33
$begingroup$
Yes sir i could not type it
$endgroup$
– user636268
Jan 21 at 16:34
$begingroup$
What does it mean for the radius of the radius of the circumcircle to be in $n$? Does it divide $n$? For a right triangle the radius of the circumcircle is half the hypotenuse.
$endgroup$
– Ross Millikan
Jan 21 at 16:54
$begingroup$
Yes, n is the product of all the given primes
$endgroup$
– user636268
Jan 21 at 16:56
1
$begingroup$
So if we let $r= 2n/13,$ for example, we can show that the right triangle with legs $5r$ and $12r$ is beautiful. So I think it comes down to the number (modulo $100$) of primitive Pythagorean triples in which the hypotenuse is a product of some subset of the first $2017$ odd primes.
$endgroup$
– David K
Jan 21 at 17:26
1
1
$begingroup$
"Let p₁,p₂,...,p₂₀₁₇, be the 2017 odd prime numbers and none are equal." -- Which primes, because there are infinitely many primes, almost all odd except $2$? Similarly -- "Let n=product of all the primes" -- this product is infinite. Is this referring to the product $p_1 times ... times p_{2017}$?
$endgroup$
– Eevee Trainer
Jan 21 at 16:33
$begingroup$
"Let p₁,p₂,...,p₂₀₁₇, be the 2017 odd prime numbers and none are equal." -- Which primes, because there are infinitely many primes, almost all odd except $2$? Similarly -- "Let n=product of all the primes" -- this product is infinite. Is this referring to the product $p_1 times ... times p_{2017}$?
$endgroup$
– Eevee Trainer
Jan 21 at 16:33
$begingroup$
Yes sir i could not type it
$endgroup$
– user636268
Jan 21 at 16:34
$begingroup$
Yes sir i could not type it
$endgroup$
– user636268
Jan 21 at 16:34
$begingroup$
What does it mean for the radius of the radius of the circumcircle to be in $n$? Does it divide $n$? For a right triangle the radius of the circumcircle is half the hypotenuse.
$endgroup$
– Ross Millikan
Jan 21 at 16:54
$begingroup$
What does it mean for the radius of the radius of the circumcircle to be in $n$? Does it divide $n$? For a right triangle the radius of the circumcircle is half the hypotenuse.
$endgroup$
– Ross Millikan
Jan 21 at 16:54
$begingroup$
Yes, n is the product of all the given primes
$endgroup$
– user636268
Jan 21 at 16:56
$begingroup$
Yes, n is the product of all the given primes
$endgroup$
– user636268
Jan 21 at 16:56
1
1
$begingroup$
So if we let $r= 2n/13,$ for example, we can show that the right triangle with legs $5r$ and $12r$ is beautiful. So I think it comes down to the number (modulo $100$) of primitive Pythagorean triples in which the hypotenuse is a product of some subset of the first $2017$ odd primes.
$endgroup$
– David K
Jan 21 at 17:26
$begingroup$
So if we let $r= 2n/13,$ for example, we can show that the right triangle with legs $5r$ and $12r$ is beautiful. So I think it comes down to the number (modulo $100$) of primitive Pythagorean triples in which the hypotenuse is a product of some subset of the first $2017$ odd primes.
$endgroup$
– David K
Jan 21 at 17:26
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
We have that:
$$n=p_1p_2...p_{2017}$$
and:
$$a^2+b^2=N=(2n)^2=2^2p_1^2p_2^2dots p_{2017}^2tag{1}$$
Essentially, on the right side you have a product of the first 2018 prime numbers squared.
Theorem 4.4 from paper Sum of two squares (Jahnavi Bhaskar) states the following (in a slightly different form):
Let $n$ be a positive integer with the following prime factorization:
$$n=2^c(p_1^{s_1}p_2^{s_2}...p_k^{s_k})(q_1^{t_1}q_2^{t_2}...q_l^{t_l})$$
...where
$$p_iequiv1pmod 4$$ $$q_iequiv3pmod 4$$ If any $t_i$ is odd, number $n$ cannot be written as the sum of two squares. If all $t_i$ are even, the number of representations of $n$ as the sum of two squares is equal to:
$$r(n)=4(s_1+1)(s_2+1)dots(s_k+1)$$
(The proof seems to be highly technical and Bhaskar provides only a reference to it)
In our problem $r(N)$ represents the number of "beautiful triangles". Based on (1) we see that all primes have the same power: $s_i=2$, $t_i=2$ and therefore the number of beautiful triangles is:
$$r(N)= 4times(2+1)^m=4times 3^m$$
...where $m$ stands for the number of odd primes equal to 1 modulo 4 among the first 2017 odd primes. You can find $m$ by computer and I doubt that there is an easier way to do it:
Length[Select[Table[Prime[i], {i, 1, 2018}], Mod[#, 4] == 1 &]]
This returns 994. So the result is:
$$r(N)=4times 3^{994}$$
Finding the last two digits is a fairly straightforward exercise from modular arithmetics:
$$r(N)equiv 76 pmod {100}$$
In my opinion, it's unlikely that you can solve this problem without some help from the computer. There is no exact formula for a number of primes equal to 1 (or 3) modulo 4 among the first $n$ primes. You have to count them one by one and that's something that you cannot do by hand.
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
$endgroup$
$begingroup$
It is a great solution, but the bhaskar reference was unknown to me. Learnt something new
$endgroup$
– user636268
Jan 22 at 10:06
$begingroup$
It is really a good and helpful reference
$endgroup$
– user636268
Jan 22 at 10:07
1
$begingroup$
@Md.ShamimAkhtar You're welcome.
$endgroup$
– Oldboy
Jan 22 at 10:08
1
$begingroup$
@Md.ShamimAkhtar Actually, I did not search any specific place. I was just investigating the topic and Google helped me to find a lot of related papers (Euler had studied this subject a lot), but Bhaskar's work was the most useful. It just popped up.
$endgroup$
– Oldboy
Jan 22 at 10:13
1
$begingroup$
@Md.ShamimAkhtar Yes, I like it. And I do math just for fun :)
$endgroup$
– Oldboy
Jan 22 at 10:28
|
show 3 more comments
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
We have that:
$$n=p_1p_2...p_{2017}$$
and:
$$a^2+b^2=N=(2n)^2=2^2p_1^2p_2^2dots p_{2017}^2tag{1}$$
Essentially, on the right side you have a product of the first 2018 prime numbers squared.
Theorem 4.4 from paper Sum of two squares (Jahnavi Bhaskar) states the following (in a slightly different form):
Let $n$ be a positive integer with the following prime factorization:
$$n=2^c(p_1^{s_1}p_2^{s_2}...p_k^{s_k})(q_1^{t_1}q_2^{t_2}...q_l^{t_l})$$
...where
$$p_iequiv1pmod 4$$ $$q_iequiv3pmod 4$$ If any $t_i$ is odd, number $n$ cannot be written as the sum of two squares. If all $t_i$ are even, the number of representations of $n$ as the sum of two squares is equal to:
$$r(n)=4(s_1+1)(s_2+1)dots(s_k+1)$$
(The proof seems to be highly technical and Bhaskar provides only a reference to it)
In our problem $r(N)$ represents the number of "beautiful triangles". Based on (1) we see that all primes have the same power: $s_i=2$, $t_i=2$ and therefore the number of beautiful triangles is:
$$r(N)= 4times(2+1)^m=4times 3^m$$
...where $m$ stands for the number of odd primes equal to 1 modulo 4 among the first 2017 odd primes. You can find $m$ by computer and I doubt that there is an easier way to do it:
Length[Select[Table[Prime[i], {i, 1, 2018}], Mod[#, 4] == 1 &]]
This returns 994. So the result is:
$$r(N)=4times 3^{994}$$
Finding the last two digits is a fairly straightforward exercise from modular arithmetics:
$$r(N)equiv 76 pmod {100}$$
In my opinion, it's unlikely that you can solve this problem without some help from the computer. There is no exact formula for a number of primes equal to 1 (or 3) modulo 4 among the first $n$ primes. You have to count them one by one and that's something that you cannot do by hand.
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
$endgroup$
$begingroup$
It is a great solution, but the bhaskar reference was unknown to me. Learnt something new
$endgroup$
– user636268
Jan 22 at 10:06
$begingroup$
It is really a good and helpful reference
$endgroup$
– user636268
Jan 22 at 10:07
1
$begingroup$
@Md.ShamimAkhtar You're welcome.
$endgroup$
– Oldboy
Jan 22 at 10:08
1
$begingroup$
@Md.ShamimAkhtar Actually, I did not search any specific place. I was just investigating the topic and Google helped me to find a lot of related papers (Euler had studied this subject a lot), but Bhaskar's work was the most useful. It just popped up.
$endgroup$
– Oldboy
Jan 22 at 10:13
1
$begingroup$
@Md.ShamimAkhtar Yes, I like it. And I do math just for fun :)
$endgroup$
– Oldboy
Jan 22 at 10:28
|
show 3 more comments
$begingroup$
We have that:
$$n=p_1p_2...p_{2017}$$
and:
$$a^2+b^2=N=(2n)^2=2^2p_1^2p_2^2dots p_{2017}^2tag{1}$$
Essentially, on the right side you have a product of the first 2018 prime numbers squared.
Theorem 4.4 from paper Sum of two squares (Jahnavi Bhaskar) states the following (in a slightly different form):
Let $n$ be a positive integer with the following prime factorization:
$$n=2^c(p_1^{s_1}p_2^{s_2}...p_k^{s_k})(q_1^{t_1}q_2^{t_2}...q_l^{t_l})$$
...where
$$p_iequiv1pmod 4$$ $$q_iequiv3pmod 4$$ If any $t_i$ is odd, number $n$ cannot be written as the sum of two squares. If all $t_i$ are even, the number of representations of $n$ as the sum of two squares is equal to:
$$r(n)=4(s_1+1)(s_2+1)dots(s_k+1)$$
(The proof seems to be highly technical and Bhaskar provides only a reference to it)
In our problem $r(N)$ represents the number of "beautiful triangles". Based on (1) we see that all primes have the same power: $s_i=2$, $t_i=2$ and therefore the number of beautiful triangles is:
$$r(N)= 4times(2+1)^m=4times 3^m$$
...where $m$ stands for the number of odd primes equal to 1 modulo 4 among the first 2017 odd primes. You can find $m$ by computer and I doubt that there is an easier way to do it:
Length[Select[Table[Prime[i], {i, 1, 2018}], Mod[#, 4] == 1 &]]
This returns 994. So the result is:
$$r(N)=4times 3^{994}$$
Finding the last two digits is a fairly straightforward exercise from modular arithmetics:
$$r(N)equiv 76 pmod {100}$$
In my opinion, it's unlikely that you can solve this problem without some help from the computer. There is no exact formula for a number of primes equal to 1 (or 3) modulo 4 among the first $n$ primes. You have to count them one by one and that's something that you cannot do by hand.
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
$endgroup$
$begingroup$
It is a great solution, but the bhaskar reference was unknown to me. Learnt something new
$endgroup$
– user636268
Jan 22 at 10:06
$begingroup$
It is really a good and helpful reference
$endgroup$
– user636268
Jan 22 at 10:07
1
$begingroup$
@Md.ShamimAkhtar You're welcome.
$endgroup$
– Oldboy
Jan 22 at 10:08
1
$begingroup$
@Md.ShamimAkhtar Actually, I did not search any specific place. I was just investigating the topic and Google helped me to find a lot of related papers (Euler had studied this subject a lot), but Bhaskar's work was the most useful. It just popped up.
$endgroup$
– Oldboy
Jan 22 at 10:13
1
$begingroup$
@Md.ShamimAkhtar Yes, I like it. And I do math just for fun :)
$endgroup$
– Oldboy
Jan 22 at 10:28
|
show 3 more comments
$begingroup$
We have that:
$$n=p_1p_2...p_{2017}$$
and:
$$a^2+b^2=N=(2n)^2=2^2p_1^2p_2^2dots p_{2017}^2tag{1}$$
Essentially, on the right side you have a product of the first 2018 prime numbers squared.
Theorem 4.4 from paper Sum of two squares (Jahnavi Bhaskar) states the following (in a slightly different form):
Let $n$ be a positive integer with the following prime factorization:
$$n=2^c(p_1^{s_1}p_2^{s_2}...p_k^{s_k})(q_1^{t_1}q_2^{t_2}...q_l^{t_l})$$
...where
$$p_iequiv1pmod 4$$ $$q_iequiv3pmod 4$$ If any $t_i$ is odd, number $n$ cannot be written as the sum of two squares. If all $t_i$ are even, the number of representations of $n$ as the sum of two squares is equal to:
$$r(n)=4(s_1+1)(s_2+1)dots(s_k+1)$$
(The proof seems to be highly technical and Bhaskar provides only a reference to it)
In our problem $r(N)$ represents the number of "beautiful triangles". Based on (1) we see that all primes have the same power: $s_i=2$, $t_i=2$ and therefore the number of beautiful triangles is:
$$r(N)= 4times(2+1)^m=4times 3^m$$
...where $m$ stands for the number of odd primes equal to 1 modulo 4 among the first 2017 odd primes. You can find $m$ by computer and I doubt that there is an easier way to do it:
Length[Select[Table[Prime[i], {i, 1, 2018}], Mod[#, 4] == 1 &]]
This returns 994. So the result is:
$$r(N)=4times 3^{994}$$
Finding the last two digits is a fairly straightforward exercise from modular arithmetics:
$$r(N)equiv 76 pmod {100}$$
In my opinion, it's unlikely that you can solve this problem without some help from the computer. There is no exact formula for a number of primes equal to 1 (or 3) modulo 4 among the first $n$ primes. You have to count them one by one and that's something that you cannot do by hand.
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
$endgroup$
We have that:
$$n=p_1p_2...p_{2017}$$
and:
$$a^2+b^2=N=(2n)^2=2^2p_1^2p_2^2dots p_{2017}^2tag{1}$$
Essentially, on the right side you have a product of the first 2018 prime numbers squared.
Theorem 4.4 from paper Sum of two squares (Jahnavi Bhaskar) states the following (in a slightly different form):
Let $n$ be a positive integer with the following prime factorization:
$$n=2^c(p_1^{s_1}p_2^{s_2}...p_k^{s_k})(q_1^{t_1}q_2^{t_2}...q_l^{t_l})$$
...where
$$p_iequiv1pmod 4$$ $$q_iequiv3pmod 4$$ If any $t_i$ is odd, number $n$ cannot be written as the sum of two squares. If all $t_i$ are even, the number of representations of $n$ as the sum of two squares is equal to:
$$r(n)=4(s_1+1)(s_2+1)dots(s_k+1)$$
(The proof seems to be highly technical and Bhaskar provides only a reference to it)
In our problem $r(N)$ represents the number of "beautiful triangles". Based on (1) we see that all primes have the same power: $s_i=2$, $t_i=2$ and therefore the number of beautiful triangles is:
$$r(N)= 4times(2+1)^m=4times 3^m$$
...where $m$ stands for the number of odd primes equal to 1 modulo 4 among the first 2017 odd primes. You can find $m$ by computer and I doubt that there is an easier way to do it:
Length[Select[Table[Prime[i], {i, 1, 2018}], Mod[#, 4] == 1 &]]
This returns 994. So the result is:
$$r(N)=4times 3^{994}$$
Finding the last two digits is a fairly straightforward exercise from modular arithmetics:
$$r(N)equiv 76 pmod {100}$$
In my opinion, it's unlikely that you can solve this problem without some help from the computer. There is no exact formula for a number of primes equal to 1 (or 3) modulo 4 among the first $n$ primes. You have to count them one by one and that's something that you cannot do by hand.
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
edited Jan 21 at 21:58
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
answered Jan 21 at 21:27
OldboyOldboy
8,62711036
8,62711036
$begingroup$
It is a great solution, but the bhaskar reference was unknown to me. Learnt something new
$endgroup$
– user636268
Jan 22 at 10:06
$begingroup$
It is really a good and helpful reference
$endgroup$
– user636268
Jan 22 at 10:07
1
$begingroup$
@Md.ShamimAkhtar You're welcome.
$endgroup$
– Oldboy
Jan 22 at 10:08
1
$begingroup$
@Md.ShamimAkhtar Actually, I did not search any specific place. I was just investigating the topic and Google helped me to find a lot of related papers (Euler had studied this subject a lot), but Bhaskar's work was the most useful. It just popped up.
$endgroup$
– Oldboy
Jan 22 at 10:13
1
$begingroup$
@Md.ShamimAkhtar Yes, I like it. And I do math just for fun :)
$endgroup$
– Oldboy
Jan 22 at 10:28
|
show 3 more comments
$begingroup$
It is a great solution, but the bhaskar reference was unknown to me. Learnt something new
$endgroup$
– user636268
Jan 22 at 10:06
$begingroup$
It is really a good and helpful reference
$endgroup$
– user636268
Jan 22 at 10:07
1
$begingroup$
@Md.ShamimAkhtar You're welcome.
$endgroup$
– Oldboy
Jan 22 at 10:08
1
$begingroup$
@Md.ShamimAkhtar Actually, I did not search any specific place. I was just investigating the topic and Google helped me to find a lot of related papers (Euler had studied this subject a lot), but Bhaskar's work was the most useful. It just popped up.
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– Oldboy
Jan 22 at 10:13
1
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@Md.ShamimAkhtar Yes, I like it. And I do math just for fun :)
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– Oldboy
Jan 22 at 10:28
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It is a great solution, but the bhaskar reference was unknown to me. Learnt something new
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– user636268
Jan 22 at 10:06
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It is a great solution, but the bhaskar reference was unknown to me. Learnt something new
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– user636268
Jan 22 at 10:06
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It is really a good and helpful reference
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– user636268
Jan 22 at 10:07
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It is really a good and helpful reference
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– user636268
Jan 22 at 10:07
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1
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@Md.ShamimAkhtar You're welcome.
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– Oldboy
Jan 22 at 10:08
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@Md.ShamimAkhtar You're welcome.
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– Oldboy
Jan 22 at 10:08
1
1
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@Md.ShamimAkhtar Actually, I did not search any specific place. I was just investigating the topic and Google helped me to find a lot of related papers (Euler had studied this subject a lot), but Bhaskar's work was the most useful. It just popped up.
$endgroup$
– Oldboy
Jan 22 at 10:13
$begingroup$
@Md.ShamimAkhtar Actually, I did not search any specific place. I was just investigating the topic and Google helped me to find a lot of related papers (Euler had studied this subject a lot), but Bhaskar's work was the most useful. It just popped up.
$endgroup$
– Oldboy
Jan 22 at 10:13
1
1
$begingroup$
@Md.ShamimAkhtar Yes, I like it. And I do math just for fun :)
$endgroup$
– Oldboy
Jan 22 at 10:28
$begingroup$
@Md.ShamimAkhtar Yes, I like it. And I do math just for fun :)
$endgroup$
– Oldboy
Jan 22 at 10:28
|
show 3 more comments
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"Let p₁,p₂,...,p₂₀₁₇, be the 2017 odd prime numbers and none are equal." -- Which primes, because there are infinitely many primes, almost all odd except $2$? Similarly -- "Let n=product of all the primes" -- this product is infinite. Is this referring to the product $p_1 times ... times p_{2017}$?
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– Eevee Trainer
Jan 21 at 16:33
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Yes sir i could not type it
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– user636268
Jan 21 at 16:34
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What does it mean for the radius of the radius of the circumcircle to be in $n$? Does it divide $n$? For a right triangle the radius of the circumcircle is half the hypotenuse.
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– Ross Millikan
Jan 21 at 16:54
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Yes, n is the product of all the given primes
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– user636268
Jan 21 at 16:56
1
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So if we let $r= 2n/13,$ for example, we can show that the right triangle with legs $5r$ and $12r$ is beautiful. So I think it comes down to the number (modulo $100$) of primitive Pythagorean triples in which the hypotenuse is a product of some subset of the first $2017$ odd primes.
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– David K
Jan 21 at 17:26