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For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that $x^n + (x+1)^n = (x+2)^n$.
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For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that:
$$x^n + (x+1)^n = (x+2)^n$$
Writing it using Newton's binom, we obtain:
$$x^n = sum_i=1^n binomni cdot x^n-i cdot (2^i - 1)$$
I don't know how to continue the problem. Can you help me, please? Thanks!
binomial-coefficients
asked Jul 18 at 17:40
Iulian Oleniuc
3619
add a comment |Â
up vote
0
down vote
favorite
For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that:
$$x^n + (x+1)^n = (x+2)^n$$
Writing it using Newton's binom, we obtain:
$$x^n = sum_i=1^n binomni cdot x^n-i cdot (2^i - 1)$$
I don't know how to continue the problem. Can you help me, please? Thanks!
binomial-coefficients
asked Jul 18 at 17:40
Iulian Oleniuc
3619
2
Fermat's Last Theorem stamps the solid proof here, but there is definitely an easier method in this special case...
– Parcly Taxel
Jul 18 at 17:42
It must be $2^n≡1 pmod x$.
– Takahiro Waki
Jul 18 at 17:45
$$beginalign(x+1)^n&=(x+2)^n-x^n\&=[(x+2)-x][(x+2)^n-1+(x+2)^x-2x+...+x^n-1]\&=2[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]endalign$$ Therefore, $x$ is odd and since $n$ is also odd, the factor $[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]$ is odd. Therefore $(x+1)^n$ is divisible by $2$ but not by $4$, which cannot happen for $n>1$.
– user577471
Jul 18 at 17:56
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that:
$$x^n + (x+1)^n = (x+2)^n$$
Writing it using Newton's binom, we obtain:
$$x^n = sum_i=1^n binomni cdot x^n-i cdot (2^i - 1)$$
I don't know how to continue the problem. Can you help me, please? Thanks!
binomial-coefficients
asked Jul 18 at 17:40
Iulian Oleniuc
3619
For any odd integer $n > 2$, show that there isn't any positive integer $x$, such that:
$$x^n + (x+1)^n = (x+2)^n$$
Writing it using Newton's binom, we obtain:
$$x^n = sum_i=1^n binomni cdot x^n-i cdot (2^i - 1)$$
I don't know how to continue the problem. Can you help me, please? Thanks!
binomial-coefficients
asked Jul 18 at 17:40
Iulian Oleniuc
3619
asked Jul 18 at 17:40
Iulian Oleniuc
3619
asked Jul 18 at 17:40
Iulian Oleniuc
3619
asked Jul 18 at 17:40
Iulian Oleniuc
3619
3619
2
Fermat's Last Theorem stamps the solid proof here, but there is definitely an easier method in this special case...
– Parcly Taxel
Jul 18 at 17:42
It must be $2^n≡1 pmod x$.
– Takahiro Waki
Jul 18 at 17:45
$$beginalign(x+1)^n&=(x+2)^n-x^n\&=[(x+2)-x][(x+2)^n-1+(x+2)^x-2x+...+x^n-1]\&=2[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]endalign$$ Therefore, $x$ is odd and since $n$ is also odd, the factor $[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]$ is odd. Therefore $(x+1)^n$ is divisible by $2$ but not by $4$, which cannot happen for $n>1$.
– user577471
Jul 18 at 17:56
add a comment |Â
2
Fermat's Last Theorem stamps the solid proof here, but there is definitely an easier method in this special case...
– Parcly Taxel
Jul 18 at 17:42
It must be $2^n≡1 pmod x$.
– Takahiro Waki
Jul 18 at 17:45
$$beginalign(x+1)^n&=(x+2)^n-x^n\&=[(x+2)-x][(x+2)^n-1+(x+2)^x-2x+...+x^n-1]\&=2[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]endalign$$ Therefore, $x$ is odd and since $n$ is also odd, the factor $[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]$ is odd. Therefore $(x+1)^n$ is divisible by $2$ but not by $4$, which cannot happen for $n>1$.
– user577471
Jul 18 at 17:56
2
2
Fermat's Last Theorem stamps the solid proof here, but there is definitely an easier method in this special case...
– Parcly Taxel
Jul 18 at 17:42
Fermat's Last Theorem stamps the solid proof here, but there is definitely an easier method in this special case...
– Parcly Taxel
Jul 18 at 17:42
It must be $2^n≡1 pmod x$.
– Takahiro Waki
Jul 18 at 17:45
It must be $2^n≡1 pmod x$.
– Takahiro Waki
Jul 18 at 17:45
$$beginalign(x+1)^n&=(x+2)^n-x^n\&=[(x+2)-x][(x+2)^n-1+(x+2)^x-2x+...+x^n-1]\&=2[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]endalign$$ Therefore, $x$ is odd and since $n$ is also odd, the factor $[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]$ is odd. Therefore $(x+1)^n$ is divisible by $2$ but not by $4$, which cannot happen for $n>1$.
– user577471
Jul 18 at 17:56
$$beginalign(x+1)^n&=(x+2)^n-x^n\&=[(x+2)-x][(x+2)^n-1+(x+2)^x-2x+...+x^n-1]\&=2[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]endalign$$ Therefore, $x$ is odd and since $n$ is also odd, the factor $[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]$ is odd. Therefore $(x+1)^n$ is divisible by $2$ but not by $4$, which cannot happen for $n>1$.
– user577471
Jul 18 at 17:56
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Claim 1: If $x$ is even then the left side is odd but right side is even so no even solution.
Claim 2: If $x$ is odd, then $x equiv pm 1 pmod4$. In which case (with $n$ being odd) we have left side is either $1+2^n pmod4$ or $-1 pmod4$, whereas the right side is $-1 pmod4$ or $1 pmod4$ respectively. For $n>2$ the two sides are not equal. Hence no solution.
answered Jul 18 at 17:58
Anurag A
22.3k12243
add a comment |Â
up vote
2
down vote
Hint: Consider the cases $x=3k$, $x=3k+1$, $x=3k-1$ separately. You should find the remainders on division by $3$ are incompatible in all three cases.
answered Jul 18 at 17:47
Chappers
55k74190
For $x=3k+1$, we have $x^n equiv (3k+1)^nequiv 1pmod3$, $(x+1)^n equiv (3k+2)^nequiv 2pmod3$, and $(x+2)^nequiv(3k+3)^nequiv 0equiv 1+2 pmod3$ for odd $n$.
– Math Lover
Jul 18 at 17:59
add a comment |Â
up vote
0
down vote
The reminders by $x$ is equal, then 　
$2^n=kx+1$
$LHD=x^n+(x+1)^n=x*(A(x)+n)+1$
$RHD=(x+2)^n=x(B(x)+2n)+kx+1=x(B(x)+2n+k)+1$
Since $n≠2n+k$, there doesn't such integers.
answered Jul 18 at 18:00
Takahiro Waki
1,974520
This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$.
– Batominovski
Jul 29 at 12:28
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Claim 1: If $x$ is even then the left side is odd but right side is even so no even solution.
Claim 2: If $x$ is odd, then $x equiv pm 1 pmod4$. In which case (with $n$ being odd) we have left side is either $1+2^n pmod4$ or $-1 pmod4$, whereas the right side is $-1 pmod4$ or $1 pmod4$ respectively. For $n>2$ the two sides are not equal. Hence no solution.
answered Jul 18 at 17:58
Anurag A
22.3k12243
add a comment |Â
up vote
1
down vote
accepted
Claim 1: If $x$ is even then the left side is odd but right side is even so no even solution.
Claim 2: If $x$ is odd, then $x equiv pm 1 pmod4$. In which case (with $n$ being odd) we have left side is either $1+2^n pmod4$ or $-1 pmod4$, whereas the right side is $-1 pmod4$ or $1 pmod4$ respectively. For $n>2$ the two sides are not equal. Hence no solution.
answered Jul 18 at 17:58
Anurag A
22.3k12243
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Claim 1: If $x$ is even then the left side is odd but right side is even so no even solution.
Claim 2: If $x$ is odd, then $x equiv pm 1 pmod4$. In which case (with $n$ being odd) we have left side is either $1+2^n pmod4$ or $-1 pmod4$, whereas the right side is $-1 pmod4$ or $1 pmod4$ respectively. For $n>2$ the two sides are not equal. Hence no solution.
answered Jul 18 at 17:58
Anurag A
22.3k12243
Claim 1: If $x$ is even then the left side is odd but right side is even so no even solution.
Claim 2: If $x$ is odd, then $x equiv pm 1 pmod4$. In which case (with $n$ being odd) we have left side is either $1+2^n pmod4$ or $-1 pmod4$, whereas the right side is $-1 pmod4$ or $1 pmod4$ respectively. For $n>2$ the two sides are not equal. Hence no solution.
answered Jul 18 at 17:58
Anurag A
22.3k12243
answered Jul 18 at 17:58
Anurag A
22.3k12243
answered Jul 18 at 17:58
Anurag A
22.3k12243
answered Jul 18 at 17:58
Anurag A
22.3k12243
22.3k12243
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: Consider the cases $x=3k$, $x=3k+1$, $x=3k-1$ separately. You should find the remainders on division by $3$ are incompatible in all three cases.
answered Jul 18 at 17:47
Chappers
55k74190
For $x=3k+1$, we have $x^n equiv (3k+1)^nequiv 1pmod3$, $(x+1)^n equiv (3k+2)^nequiv 2pmod3$, and $(x+2)^nequiv(3k+3)^nequiv 0equiv 1+2 pmod3$ for odd $n$.
– Math Lover
Jul 18 at 17:59
add a comment |Â
up vote
2
down vote
Hint: Consider the cases $x=3k$, $x=3k+1$, $x=3k-1$ separately. You should find the remainders on division by $3$ are incompatible in all three cases.
answered Jul 18 at 17:47
Chappers
55k74190
For $x=3k+1$, we have $x^n equiv (3k+1)^nequiv 1pmod3$, $(x+1)^n equiv (3k+2)^nequiv 2pmod3$, and $(x+2)^nequiv(3k+3)^nequiv 0equiv 1+2 pmod3$ for odd $n$.
– Math Lover
Jul 18 at 17:59
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Consider the cases $x=3k$, $x=3k+1$, $x=3k-1$ separately. You should find the remainders on division by $3$ are incompatible in all three cases.
answered Jul 18 at 17:47
Chappers
55k74190
Hint: Consider the cases $x=3k$, $x=3k+1$, $x=3k-1$ separately. You should find the remainders on division by $3$ are incompatible in all three cases.
answered Jul 18 at 17:47
Chappers
55k74190
answered Jul 18 at 17:47
Chappers
55k74190
answered Jul 18 at 17:47
Chappers
55k74190
answered Jul 18 at 17:47
Chappers
55k74190
55k74190
For $x=3k+1$, we have $x^n equiv (3k+1)^nequiv 1pmod3$, $(x+1)^n equiv (3k+2)^nequiv 2pmod3$, and $(x+2)^nequiv(3k+3)^nequiv 0equiv 1+2 pmod3$ for odd $n$.
– Math Lover
Jul 18 at 17:59
add a comment |Â
For $x=3k+1$, we have $x^n equiv (3k+1)^nequiv 1pmod3$, $(x+1)^n equiv (3k+2)^nequiv 2pmod3$, and $(x+2)^nequiv(3k+3)^nequiv 0equiv 1+2 pmod3$ for odd $n$.
– Math Lover
Jul 18 at 17:59
For $x=3k+1$, we have $x^n equiv (3k+1)^nequiv 1pmod3$, $(x+1)^n equiv (3k+2)^nequiv 2pmod3$, and $(x+2)^nequiv(3k+3)^nequiv 0equiv 1+2 pmod3$ for odd $n$.
– Math Lover
Jul 18 at 17:59
For $x=3k+1$, we have $x^n equiv (3k+1)^nequiv 1pmod3$, $(x+1)^n equiv (3k+2)^nequiv 2pmod3$, and $(x+2)^nequiv(3k+3)^nequiv 0equiv 1+2 pmod3$ for odd $n$.
– Math Lover
Jul 18 at 17:59
add a comment |Â
up vote
0
down vote
The reminders by $x$ is equal, then 　
$2^n=kx+1$
$LHD=x^n+(x+1)^n=x*(A(x)+n)+1$
$RHD=(x+2)^n=x(B(x)+2n)+kx+1=x(B(x)+2n+k)+1$
Since $n≠2n+k$, there doesn't such integers.
answered Jul 18 at 18:00
Takahiro Waki
1,974520
This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$.
– Batominovski
Jul 29 at 12:28
add a comment |Â
up vote
0
down vote
The reminders by $x$ is equal, then 　
$2^n=kx+1$
$LHD=x^n+(x+1)^n=x*(A(x)+n)+1$
$RHD=(x+2)^n=x(B(x)+2n)+kx+1=x(B(x)+2n+k)+1$
Since $n≠2n+k$, there doesn't such integers.
answered Jul 18 at 18:00
Takahiro Waki
1,974520
This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$.
– Batominovski
Jul 29 at 12:28
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The reminders by $x$ is equal, then 　
$2^n=kx+1$
$LHD=x^n+(x+1)^n=x*(A(x)+n)+1$
$RHD=(x+2)^n=x(B(x)+2n)+kx+1=x(B(x)+2n+k)+1$
Since $n≠2n+k$, there doesn't such integers.
answered Jul 18 at 18:00
Takahiro Waki
1,974520
The reminders by $x$ is equal, then 　
$2^n=kx+1$
$LHD=x^n+(x+1)^n=x*(A(x)+n)+1$
$RHD=(x+2)^n=x(B(x)+2n)+kx+1=x(B(x)+2n+k)+1$
Since $n≠2n+k$, there doesn't such integers.
answered Jul 18 at 18:00
Takahiro Waki
1,974520
answered Jul 18 at 18:00
Takahiro Waki
1,974520
answered Jul 18 at 18:00
Takahiro Waki
1,974520
answered Jul 18 at 18:00
Takahiro Waki
1,974520
1,974520
This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$.
– Batominovski
Jul 29 at 12:28
add a comment |Â
This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$.
– Batominovski
Jul 29 at 12:28
This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$.
– Batominovski
Jul 29 at 12:28
This is one answer that makes no sense. Division by a polynomial variable $x$ is very different from division by a specific integer $x$.
– Batominovski
Jul 29 at 12:28
add a comment |Â
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2
Fermat's Last Theorem stamps the solid proof here, but there is definitely an easier method in this special case...
– Parcly Taxel
Jul 18 at 17:42
It must be $2^n≡1 pmod x$.
– Takahiro Waki
Jul 18 at 17:45
$$beginalign(x+1)^n&=(x+2)^n-x^n\&=[(x+2)-x][(x+2)^n-1+(x+2)^x-2x+...+x^n-1]\&=2[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]endalign$$ Therefore, $x$ is odd and since $n$ is also odd, the factor $[(x+2)^n-1+(x+2)^x-2x+...+x^n-1]$ is odd. Therefore $(x+1)^n$ is divisible by $2$ but not by $4$, which cannot happen for $n>1$.
– user577471
Jul 18 at 17:56