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Two years ago, I made a conjecture [on stackexchange]:

Today, I tried to find all solutions in integers $a,b,c$ to $$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$

I have found some solutions, such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$

Here is another simple solution: $$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My question is: are there solutions of another form (or have we found all solutions)? [on stackexchange]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803https://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

Two years ago, I made a conjecture [on stackexchange]:

Today, I tried to find all solutions in integers $a,b,c$ to $$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$

I have found some solutions, such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$

Here is another simple solution: $$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My question is: are there solutions of another form (or have we found all solutions)? [on stackexchange]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

Two years ago, I made a conjecture [on stackexchange]:

Today, I tried to find all solutions in integers $a,b,c$ to $$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$

I have found some solutions, such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$

Here is another simple solution: $$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My question is: are there solutions of another form (or have we found all solutions)? [on stackexchange]:https://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

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James Cranch
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twoTwo years ago,I have conjectures this problem I made a conjecture [on stackexchange]:[stack]

todayToday,I try Find I tried to find all solutions in integers $a,b,c$ to $$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$

$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
. Now I have found some solutionsolutions,such such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}),p>2+\sqrt{3},p\in\mathbb {Q}^{+}$$$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$

Add other simple solution Here is another simple solution: $$(a,b,c)=(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p})$$$$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My Questionquestion is: Have otherare there solutions of another form solution(or findhave we found all solutions?)? [stack][on stackexchange]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

two years ago,I have conjectures this problem:[stack]

today,I try Find all solutions $a,b,c$

$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
. Now I found some solution,such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}),p>2+\sqrt{3},p\in\mathbb {Q}^{+}$$

Add other simple solution $$(a,b,c)=(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p})$$

My Question: Have other form solution(or find all solutions?)? [stack]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

Two years ago, I made a conjecture [on stackexchange]:

Today, I tried to find all solutions in integers $a,b,c$ to $$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$

I have found some solutions, such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=\left(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}\right),\quad\text{for $p>2+\sqrt{3}$ and $p\in\mathbb {Q}^{+}$}.$$

Here is another simple solution: $$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$

My question is: are there solutions of another form (or have we found all solutions)? [on stackexchange]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

added 120 characters in body
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math110
math110
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two years ago,I have conjectures this problem:[stack]

today,I try Find all solutions $a,b,c$

$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
. Now I found some solution,such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}),p>2+\sqrt{3},p\in\mathbb {Q}^{+}$$

Add other simple solution $$(a,b,c)=(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p})$$

My Question: Have other form solution(or find all solutions?)? [stack]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

two years ago,I have conjectures this problem:[stack]

today,I try Find all solutions $a,b,c$

$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
. Now I found some solution,such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}),p>2+\sqrt{3},p\in\mathbb {Q}^{+}$$

My Question: Have other form solution(or find all solutions?)? [stack]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

two years ago,I have conjectures this problem:[stack]

today,I try Find all solutions $a,b,c$

$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
. Now I found some solution,such as $$(a,b,c)=(5/17,1/11,8/9),(1/7,5/16,9/11),(3/4,11/21,1/10),\cdots$$

$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}),p>2+\sqrt{3},p\in\mathbb {Q}^{+}$$

Add other simple solution $$(a,b,c)=(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p})$$

My Question: Have other form solution(or find all solutions?)? [stack]:http://math.stackexchange.com/questions/334448/1-a21-b21-c2-8abc-a-b-c-in-mathbbq-has-infinitely-many-sol/336803#336803

Post Reopened by Stefan Kohl♦, Joseph O'Rourke, Derek Holt, Ricardo Andrade, Joe Silverman
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math110
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Post Closed as "Not suitable for this site" by Will Jagy, Alex Degtyarev, Daniel Loughran, Stefan Kohl♦, Carlo Beenakker
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