Solve $an^2+bn+c=k^2$ over the integers.
I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.
$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$
After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.
So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?
diophantine-equations
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
add a comment |
I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.
$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$
After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.
So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?
diophantine-equations
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
math.stackexchange.com/questions/2424587/…
– individ
Jan 5 at 5:09
add a comment |
I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.
$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$
After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.
So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?
diophantine-equations
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.
$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$
After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.
So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?
diophantine-equations
diophantine-equations
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
asked Jan 4 at 20:17
SmileyCraftSmileyCraft
3,321516
3,321516
math.stackexchange.com/questions/2424587/…
– individ
Jan 5 at 5:09
add a comment |
math.stackexchange.com/questions/2424587/…
– individ
Jan 5 at 5:09
math.stackexchange.com/questions/2424587/…
– individ
Jan 5 at 5:09
math.stackexchange.com/questions/2424587/…
– individ
Jan 5 at 5:09
add a comment |
1 Answer
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Complete the square, and you get
$$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
or $$ (5n+7)^2 - 44 = 5 k^2 $$
With $x = 5n+7$, this is a Pell-type equation
$$ x^2 - 44 = 5 k^2 $$
where you only want solutions where $x equiv 2 mod 5$.
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
add a comment |
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1 Answer
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Complete the square, and you get
$$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
or $$ (5n+7)^2 - 44 = 5 k^2 $$
With $x = 5n+7$, this is a Pell-type equation
$$ x^2 - 44 = 5 k^2 $$
where you only want solutions where $x equiv 2 mod 5$.
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
add a comment |
Complete the square, and you get
$$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
or $$ (5n+7)^2 - 44 = 5 k^2 $$
With $x = 5n+7$, this is a Pell-type equation
$$ x^2 - 44 = 5 k^2 $$
where you only want solutions where $x equiv 2 mod 5$.
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
add a comment |
Complete the square, and you get
$$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
or $$ (5n+7)^2 - 44 = 5 k^2 $$
With $x = 5n+7$, this is a Pell-type equation
$$ x^2 - 44 = 5 k^2 $$
where you only want solutions where $x equiv 2 mod 5$.
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
Complete the square, and you get
$$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
or $$ (5n+7)^2 - 44 = 5 k^2 $$
With $x = 5n+7$, this is a Pell-type equation
$$ x^2 - 44 = 5 k^2 $$
where you only want solutions where $x equiv 2 mod 5$.
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
answered Jan 4 at 20:32
Robert IsraelRobert Israel
319k23208458
319k23208458
add a comment |
add a comment |
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math.stackexchange.com/questions/2424587/…
– individ
Jan 5 at 5:09