Practice Problems For Choosing mod
I am posting some problems together for the purpose(though the idea of posting no more than one problem was from myself, as far I remember).
Problem 1: Find all positive integer solutions to the equation: \[x_1^4+x_2^4+...+x_{14}^4=1599\]
Problem 2: Find primes $p$ such that $11^p+10^p$ is a perfect power.(Note: This can be done with LTE as well.)
Problem 3: Find all primes $p,q$ such that $pq+55p$ and $pq-55q$ are both squares.
Problem 4: Find all positive integers $(a,b)$ such that $2^a+3^b$ is a perfect square.
Problem 5: Find all positive integers $k$ and primes $p$ such that $p^k+1$ is a perfect square.
Problem 6: If \[x^3+y^3=z^3\] then prove that $7|xyz$.
Problem 1: Find all positive integer solutions to the equation: \[x_1^4+x_2^4+...+x_{14}^4=1599\]
Problem 2: Find primes $p$ such that $11^p+10^p$ is a perfect power.(Note: This can be done with LTE as well.)
Problem 3: Find all primes $p,q$ such that $pq+55p$ and $pq-55q$ are both squares.
Problem 4: Find all positive integers $(a,b)$ such that $2^a+3^b$ is a perfect square.
Problem 5: Find all positive integers $k$ and primes $p$ such that $p^k+1$ is a perfect square.
Problem 6: If \[x^3+y^3=z^3\] then prove that $7|xyz$.
One one thing is neutral in the universe, that is $0$.
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Re: Practice Problems For Choosing mod
Is it possible???Masum wrote:I am posting some problems together for the purpose(though the idea of posting no more than one problem was from myself, as far I remember).
Problem 1: Find all positive integer solutions to the equation: \[x_1^4+x_2^4+...+x_{14}^4=1599\]
Problem 2: Find primes $p$ such that $11^p+10^p$ is a perfect power.(Note: This can be done with LTE as well.)
Problem 3: Find all primes $p,q$ such that $pq+55p$ and $pq-55q$ are both squares.
Problem 4: Find all positive integers $(a,b)$ such that $2^a+3^b$ is a perfect square.
Problem 5: Find all positive integers $k$ and primes $p$ such that $p^k+1$ is a perfect square.
Problem 6: If \[x^3+y^3=z^3\] then prove that $7|xyz$.
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- afif mansib ch
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Re: Practice Problems For Choosing mod
i think you said about the 6th problem.
\[x,y,z=(7k+1),(7k+2)..(7k+7)\]
\[x^3,y^3,z^3\equiv 0,1,6(mod7)\]
ca be done as:\[z^3\equiv 0(mod7),x^3+y^3\equiv 0+0/1+6/6+1(mod7)\]
\[z^3\equiv 1(mod7),x^3+y^3\equiv 0+1/1+0(mod7)\]
\[z^3\equiv 6(mod7),x^3+y^3\equiv 0+6/6+0(mod7)\]
every time x,y or z =7k.so xyz is divisible by 7.
\[x,y,z=(7k+1),(7k+2)..(7k+7)\]
\[x^3,y^3,z^3\equiv 0,1,6(mod7)\]
ca be done as:\[z^3\equiv 0(mod7),x^3+y^3\equiv 0+0/1+6/6+1(mod7)\]
\[z^3\equiv 1(mod7),x^3+y^3\equiv 0+1/1+0(mod7)\]
\[z^3\equiv 6(mod7),x^3+y^3\equiv 0+6/6+0(mod7)\]
every time x,y or z =7k.so xyz is divisible by 7.
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Re: Practice Problems For Choosing mod
Can u give me a practical example where x,y and z are real numbers?
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Re: Practice Problems For Choosing mod
there is none according to the famous 'fermat's last theorem'
But, the problem says if there were any examples, $7$ would divide $xyz$
But, the problem says if there were any examples, $7$ would divide $xyz$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Practice Problems For Choosing mod
He clearly means when they are integers. Why are you even thinking about real numbers when talking about divisibility? What does $7$ divides a real number mean?sakibtanvir wrote:Can u give me a practical example where x,y and z are real numbers?
EDIT: And if you want an example for integers, here is one: $x=1,y=0,z=1$ which satisfy $x^3+y^3=z^3$, and $7\mid xyz$.
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- Tahmid Hasan
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Re: Practice Problems For Choosing mod
1.$x^4 \equiv 1,0 (mod 16)$
so $x_1^4+x_2^4+x_3^4+...+x_{14}^4 $ can have a remainder at most $14$ but $1599 \equiv 15(mod 16)$,a contradiction.
3.$pq+55p=p(q+55)$
so,$p \mid q+55$[since $p$ is prime]
so we can say $q+55=xp$[$x \geq 1$]\
in the same way we conclude $q \mid p-55$
so,$p-55=yq$[$y \geq 1,p>55$]
so,$q+55=xp=x(yq+55)$
or,$q(1-xy)=55(x-1)$
since $x-1 \geq 0,55(x-1)=q(1-xy) \geq 0$
hence $1-xy \geq 0$
so we deduce $x=1,y=1$
thus $p=q+55$,since the only even prime is $2$,so$q=2$
but then $p=57=3*19$,so a contradiction.
4.let $2^a+3^b=k^2$,since LHS is odd $k^2 \equiv 1 (mod 8)$
$2^a \equiv 2,4,0 (mod 8)$[$2$ when $a=1$,$4$ when $a=2$ and $0$ for all $a \geq 3$]
$3^b \equiv 1,3 (mod 8)$[$1$ when $b$ is even and $3$ when $b$ is odd]
from here we conclude $b=2t$,[$t$ is natural]
so,$2^a=(k+3^t)(k-3^t)$
hence $k+3^t=2^m,k-3^t=2^{a-m}$[$m>a-m$]
so,$2*3^t=2^{a-m}(2^{2m-a}-1)$
so me can deduce $a=m+1$,so $3^t=2^{1-m}-1$,which is a contradiction sine $m,t$ are both natural.
5.let $p^k+1=t^2$
let $p=2$,then $2^k=(t+1)(t-1)$
let $t+1=2^a,t-1=2^b$
from here we deduce $2^b+2=2^a$so,$b=1,a=2$
so,$(p,k)=(2,3)$
if $p$ is odd then $p^k=(t+1)(t-1)$,$(t+1,t-1)=1$ since both are odd
so a contradiction.
6. has already been proved and my solution is quite the same.
and i'm still working on 2. :'(
so $x_1^4+x_2^4+x_3^4+...+x_{14}^4 $ can have a remainder at most $14$ but $1599 \equiv 15(mod 16)$,a contradiction.
3.$pq+55p=p(q+55)$
so,$p \mid q+55$[since $p$ is prime]
so we can say $q+55=xp$[$x \geq 1$]\
in the same way we conclude $q \mid p-55$
so,$p-55=yq$[$y \geq 1,p>55$]
so,$q+55=xp=x(yq+55)$
or,$q(1-xy)=55(x-1)$
since $x-1 \geq 0,55(x-1)=q(1-xy) \geq 0$
hence $1-xy \geq 0$
so we deduce $x=1,y=1$
thus $p=q+55$,since the only even prime is $2$,so$q=2$
but then $p=57=3*19$,so a contradiction.
4.let $2^a+3^b=k^2$,since LHS is odd $k^2 \equiv 1 (mod 8)$
$2^a \equiv 2,4,0 (mod 8)$[$2$ when $a=1$,$4$ when $a=2$ and $0$ for all $a \geq 3$]
$3^b \equiv 1,3 (mod 8)$[$1$ when $b$ is even and $3$ when $b$ is odd]
from here we conclude $b=2t$,[$t$ is natural]
so,$2^a=(k+3^t)(k-3^t)$
hence $k+3^t=2^m,k-3^t=2^{a-m}$[$m>a-m$]
so,$2*3^t=2^{a-m}(2^{2m-a}-1)$
so me can deduce $a=m+1$,so $3^t=2^{1-m}-1$,which is a contradiction sine $m,t$ are both natural.
5.let $p^k+1=t^2$
let $p=2$,then $2^k=(t+1)(t-1)$
let $t+1=2^a,t-1=2^b$
from here we deduce $2^b+2=2^a$so,$b=1,a=2$
so,$(p,k)=(2,3)$
if $p$ is odd then $p^k=(t+1)(t-1)$,$(t+1,t-1)=1$ since both are odd
so a contradiction.
6. has already been proved and my solution is quite the same.
and i'm still working on 2. :'(
Last edited by Masum on Sat Feb 04, 2012 3:45 pm, edited 2 times in total.
Reason: typo may be or silly mistake
Reason: typo may be or silly mistake
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- Nadim Ul Abrar
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Re: Practice Problems For Choosing mod
problem 2:
Does perfect power of $a$ mean $a^b$, where $b>1$?
If that is, then
$3|10^p+11^p$
so that at least $9|10^p+11^p$
Now $10^p+11^p\equiv1+2^p=0(mod9)$
that leads$ 2^{2p} \equiv1 (mod9)$
again $2^6 \equiv 1mod9$
so$ 2^{gcd(6,2p)}=1 mod9$
it leads $p=3$ cz $2^2\equiv 4mod9$ , so $3|p$ or $p=3$
Now $7|10^3+11^3 $, but$ 7^2$ doesn't divide $10^3+11^3 $
(with LTE we can easily show that the max power of $3$ in $10^p+11^p$ is $1$ , if $p\neq3$ .
If $p=3$then its $2$ )
Does perfect power of $a$ mean $a^b$, where $b>1$?
If that is, then
$3|10^p+11^p$
so that at least $9|10^p+11^p$
Now $10^p+11^p\equiv1+2^p=0(mod9)$
that leads$ 2^{2p} \equiv1 (mod9)$
again $2^6 \equiv 1mod9$
so$ 2^{gcd(6,2p)}=1 mod9$
it leads $p=3$ cz $2^2\equiv 4mod9$ , so $3|p$ or $p=3$
Now $7|10^3+11^3 $, but$ 7^2$ doesn't divide $10^3+11^3 $
(with LTE we can easily show that the max power of $3$ in $10^p+11^p$ is $1$ , if $p\neq3$ .
If $p=3$then its $2$ )
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Re: Practice Problems For Choosing mod
Taking $mod$ is a standard way. But it seems you know LTE. Then you can make your proof much simpler. LTE is used to make our jobs much easier(and so is Zsigmondy's theorem).
If $11^p+10^p$ is a perfect power, then it is divisible by $3$, to be a power it then must be divisible by $9$. But from LTE, \[\nu_3(11^p+10^p)=\nu_3(11+10)+\nu_3(p)\le2\]
Equality occurs if $p=3$. But it is not possible by easy check. $p=2$ does not fit as well.
If $11^p+10^p$ is a perfect power, then it is divisible by $3$, to be a power it then must be divisible by $9$. But from LTE, \[\nu_3(11^p+10^p)=\nu_3(11+10)+\nu_3(p)\le2\]
Equality occurs if $p=3$. But it is not possible by easy check. $p=2$ does not fit as well.
One one thing is neutral in the universe, that is $0$.
- FahimFerdous
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Re: Practice Problems For Choosing mod
Tahmid, your solution to problem 1 must be edited. The process is correct, but some editing must be done.
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