About Fermat numbers
$begingroup$
I am self studying elementary number theory by David Burton and got stuck on this problem in exercise $11.4$ of chapter - numbers of special form.
Question is:
for $n> 0$ prove that $F_n$ ( Fermat number $n$) is never a triangular number.
I tried to attempt it by assuming it to be triangular and obtaining contradiction in quadratic equation thus formed but was struck.
Please help.
elementary-number-theory
edited Dec 2 '18 at 19:29
Bernard
119k639112
asked Dec 2 '18 at 18:15
AMDEAMDE
62
$endgroup$
add a comment |
$begingroup$
I am self studying elementary number theory by David Burton and got stuck on this problem in exercise $11.4$ of chapter - numbers of special form.
Question is:
for $n> 0$ prove that $F_n$ ( Fermat number $n$) is never a triangular number.
I tried to attempt it by assuming it to be triangular and obtaining contradiction in quadratic equation thus formed but was struck.
Please help.
elementary-number-theory
edited Dec 2 '18 at 19:29
Bernard
119k639112
asked Dec 2 '18 at 18:15
AMDEAMDE
62
$endgroup$
add a comment |
$begingroup$
I am self studying elementary number theory by David Burton and got stuck on this problem in exercise $11.4$ of chapter - numbers of special form.
Question is:
for $n> 0$ prove that $F_n$ ( Fermat number $n$) is never a triangular number.
I tried to attempt it by assuming it to be triangular and obtaining contradiction in quadratic equation thus formed but was struck.
Please help.
elementary-number-theory
edited Dec 2 '18 at 19:29
Bernard
119k639112
asked Dec 2 '18 at 18:15
AMDEAMDE
62
$endgroup$
I am self studying elementary number theory by David Burton and got stuck on this problem in exercise $11.4$ of chapter - numbers of special form.
Question is:
for $n> 0$ prove that $F_n$ ( Fermat number $n$) is never a triangular number.
I tried to attempt it by assuming it to be triangular and obtaining contradiction in quadratic equation thus formed but was struck.
Please help.
elementary-number-theory
elementary-number-theory
edited Dec 2 '18 at 19:29
Bernard
119k639112
asked Dec 2 '18 at 18:15
AMDEAMDE
62
edited Dec 2 '18 at 19:29
Bernard
119k639112
asked Dec 2 '18 at 18:15
AMDEAMDE
62
edited Dec 2 '18 at 19:29
Bernard
119k639112
edited Dec 2 '18 at 19:29
Bernard
119k639112
edited Dec 2 '18 at 19:29
Bernard
119k639112
119k639112
asked Dec 2 '18 at 18:15
AMDEAMDE
62
asked Dec 2 '18 at 18:15
AMDEAMDE
62
asked Dec 2 '18 at 18:15
AMDEAMDE
62
62
add a comment |
add a comment |
1 Answer
1
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votes
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Let's just try to solve it directly. We want $$2^{2^n}+1=frac {k(k+1)}2implies k^2+k-2(2^{2^n}+1)=0$$
For this to have an integer solution, or even a rational one, we would need the discriminant to be the square of an integer. Thus we require that $$sqrt {1+8(2^{2^n}+1)}in mathbb N$$
Thus we want $$1+8(2^{2^n}+1)=m^2implies 2^{2^n+3}=m^2-9=(m+3)(m-3)$$
Now, this is nearly impossible. To achieve it, we'd need to have both $m-3,m+3$ powers of $2$ but since $m-3, m+3$ differ by $6$ the only solutions would be very small. Indeed the only powers of $2$ that differ by $6$ are $(2,8)$. It is easy to show that $n=0, k=2,m=5$ is the only small solution and the above shows that there are no large ones, so we are done.
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
$endgroup$
$begingroup$
Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that.
$endgroup$
– AMDE
Dec 3 '18 at 3:18
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Let's just try to solve it directly. We want $$2^{2^n}+1=frac {k(k+1)}2implies k^2+k-2(2^{2^n}+1)=0$$
For this to have an integer solution, or even a rational one, we would need the discriminant to be the square of an integer. Thus we require that $$sqrt {1+8(2^{2^n}+1)}in mathbb N$$
Thus we want $$1+8(2^{2^n}+1)=m^2implies 2^{2^n+3}=m^2-9=(m+3)(m-3)$$
Now, this is nearly impossible. To achieve it, we'd need to have both $m-3,m+3$ powers of $2$ but since $m-3, m+3$ differ by $6$ the only solutions would be very small. Indeed the only powers of $2$ that differ by $6$ are $(2,8)$. It is easy to show that $n=0, k=2,m=5$ is the only small solution and the above shows that there are no large ones, so we are done.
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
$endgroup$
$begingroup$
Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that.
$endgroup$
– AMDE
Dec 3 '18 at 3:18
add a comment |
$begingroup$
Let's just try to solve it directly. We want $$2^{2^n}+1=frac {k(k+1)}2implies k^2+k-2(2^{2^n}+1)=0$$
For this to have an integer solution, or even a rational one, we would need the discriminant to be the square of an integer. Thus we require that $$sqrt {1+8(2^{2^n}+1)}in mathbb N$$
Thus we want $$1+8(2^{2^n}+1)=m^2implies 2^{2^n+3}=m^2-9=(m+3)(m-3)$$
Now, this is nearly impossible. To achieve it, we'd need to have both $m-3,m+3$ powers of $2$ but since $m-3, m+3$ differ by $6$ the only solutions would be very small. Indeed the only powers of $2$ that differ by $6$ are $(2,8)$. It is easy to show that $n=0, k=2,m=5$ is the only small solution and the above shows that there are no large ones, so we are done.
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
$endgroup$
$begingroup$
Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that.
$endgroup$
– AMDE
Dec 3 '18 at 3:18
add a comment |
$begingroup$
Let's just try to solve it directly. We want $$2^{2^n}+1=frac {k(k+1)}2implies k^2+k-2(2^{2^n}+1)=0$$
For this to have an integer solution, or even a rational one, we would need the discriminant to be the square of an integer. Thus we require that $$sqrt {1+8(2^{2^n}+1)}in mathbb N$$
Thus we want $$1+8(2^{2^n}+1)=m^2implies 2^{2^n+3}=m^2-9=(m+3)(m-3)$$
Now, this is nearly impossible. To achieve it, we'd need to have both $m-3,m+3$ powers of $2$ but since $m-3, m+3$ differ by $6$ the only solutions would be very small. Indeed the only powers of $2$ that differ by $6$ are $(2,8)$. It is easy to show that $n=0, k=2,m=5$ is the only small solution and the above shows that there are no large ones, so we are done.
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
$endgroup$
Let's just try to solve it directly. We want $$2^{2^n}+1=frac {k(k+1)}2implies k^2+k-2(2^{2^n}+1)=0$$
For this to have an integer solution, or even a rational one, we would need the discriminant to be the square of an integer. Thus we require that $$sqrt {1+8(2^{2^n}+1)}in mathbb N$$
Thus we want $$1+8(2^{2^n}+1)=m^2implies 2^{2^n+3}=m^2-9=(m+3)(m-3)$$
Now, this is nearly impossible. To achieve it, we'd need to have both $m-3,m+3$ powers of $2$ but since $m-3, m+3$ differ by $6$ the only solutions would be very small. Indeed the only powers of $2$ that differ by $6$ are $(2,8)$. It is easy to show that $n=0, k=2,m=5$ is the only small solution and the above shows that there are no large ones, so we are done.
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
edited Dec 2 '18 at 18:55
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
answered Dec 2 '18 at 18:48
lulululu
39.4k24677
39.4k24677
$begingroup$
Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that.
$endgroup$
– AMDE
Dec 3 '18 at 3:18
add a comment |
$begingroup$
Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that.
$endgroup$
– AMDE
Dec 3 '18 at 3:18
$begingroup$
Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that.
$endgroup$
– AMDE
Dec 3 '18 at 3:18
$begingroup$
Thank you very much for your answer. I reached till factoring in m-3 and m+3 but could not think beyond that.
$endgroup$
– AMDE
Dec 3 '18 at 3:18
add a comment |
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