Show that the only divisors are $1$ and $2$ for $(z^(2^x)+1)$ and $(z^(2^y)+1)$ where $x,y,zinmathbbN$Fermat numbers are coprimeHow to deduce that $gcd(a, b) = gcd(a, a+b)$ given that the common divisors of $a$ and $b$ are the same as the common divisors of $a$ and $a+b$?How to show that if $24k+4$ and $24k+1$ are both perfect squares where $k$ is a natural number, then it is only when $k=0$?Show that the odd prime divisors of $n^2+1$ are of form $4k+1$Show that $1$ and $-1$ are the only divisors of $1$If $p$ is prime and $gcd(m,p) = 1$ show that $gcd(m,p^k) = 1$Show that $x^2 + y^2$ and $x^2 - y^2$ cannot both be perfect squares at the same time where $x, y in mathbbZ^+$.Find the common divisors of $a_1986$ and $a_6891$numbers that are the difference of two integers that have an odd number of divisorsIf $d_1 =1, d_2, cdots, d_k = n$ are positive divisors of natural $n$, show $(d_1d_2 cdots d_k)^2 = n^k$.Show that only finitely many positive integers $n$ have the property that for all $m$ for which $1<m<n$ and $textgcd(m,n) = 1$, $m$ is prime.

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Show that the only divisors are $1$ and $2$ for $(z^(2^x)+1)$ and $(z^(2^y)+1)$ where $x,y,zinmathbbN$


Fermat numbers are coprimeHow to deduce that $gcd(a, b) = gcd(a, a+b)$ given that the common divisors of $a$ and $b$ are the same as the common divisors of $a$ and $a+b$?How to show that if $24k+4$ and $24k+1$ are both perfect squares where $k$ is a natural number, then it is only when $k=0$?Show that the odd prime divisors of $n^2+1$ are of form $4k+1$Show that $1$ and $-1$ are the only divisors of $1$If $p$ is prime and $gcd(m,p) = 1$ show that $gcd(m,p^k) = 1$Show that $x^2 + y^2$ and $x^2 - y^2$ cannot both be perfect squares at the same time where $x, y in mathbbZ^+$.Find the common divisors of $a_1986$ and $a_6891$numbers that are the difference of two integers that have an odd number of divisorsIf $d_1 =1, d_2, cdots, d_k = n$ are positive divisors of natural $n$, show $(d_1d_2 cdots d_k)^2 = n^k$.Show that only finitely many positive integers $n$ have the property that for all $m$ for which $1<m<n$ and $textgcd(m,n) = 1$, $m$ is prime.













0












$begingroup$


I am trying to show that the only divisors are $1$ and $2$ for both $(z^(2^x)+1)$ and $(z^(2^y)+1)$ where $x,y,zinmathbbN$. To start the problem, the logical choice is to use difference of squares. We see that $z^2^x+1-(z^2^y+1)=z^2^x-z^2^y$. I am not sure where to go from here except to show that the gcd is 2, which proves the result.






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    math.stackexchange.com/questions/123524/…
    $endgroup$
    – lab bhattacharjee
    Mar 14 at 4:35






  • 1




    $begingroup$
    I assume that you also want that $x neq y$, else the gcd of the $2$ values will be the values them.
    $endgroup$
    – John Omielan
    Mar 14 at 4:37















0












$begingroup$


I am trying to show that the only divisors are $1$ and $2$ for both $(z^(2^x)+1)$ and $(z^(2^y)+1)$ where $x,y,zinmathbbN$. To start the problem, the logical choice is to use difference of squares. We see that $z^2^x+1-(z^2^y+1)=z^2^x-z^2^y$. I am not sure where to go from here except to show that the gcd is 2, which proves the result.






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    math.stackexchange.com/questions/123524/…
    $endgroup$
    – lab bhattacharjee
    Mar 14 at 4:35






  • 1




    $begingroup$
    I assume that you also want that $x neq y$, else the gcd of the $2$ values will be the values them.
    $endgroup$
    – John Omielan
    Mar 14 at 4:37













0












0








0





$begingroup$


I am trying to show that the only divisors are $1$ and $2$ for both $(z^(2^x)+1)$ and $(z^(2^y)+1)$ where $x,y,zinmathbbN$. To start the problem, the logical choice is to use difference of squares. We see that $z^2^x+1-(z^2^y+1)=z^2^x-z^2^y$. I am not sure where to go from here except to show that the gcd is 2, which proves the result.






number-theory elementary-number-theory







share|cite|improve this question











$endgroup$




I am trying to show that the only divisors are $1$ and $2$ for both $(z^(2^x)+1)$ and $(z^(2^y)+1)$ where $x,y,zinmathbbN$. To start the problem, the logical choice is to use difference of squares. We see that $z^2^x+1-(z^2^y+1)=z^2^x-z^2^y$. I am not sure where to go from here except to show that the gcd is 2, which proves the result.






number-theory elementary-number-theory




number-theory elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 5:14









Robert Howard

2,2293935




2,2293935










asked Mar 14 at 4:26









user593295user593295

678




678







  • 3




    $begingroup$
    math.stackexchange.com/questions/123524/…
    $endgroup$
    – lab bhattacharjee
    Mar 14 at 4:35






  • 1




    $begingroup$
    I assume that you also want that $x neq y$, else the gcd of the $2$ values will be the values them.
    $endgroup$
    – John Omielan
    Mar 14 at 4:37












  • 3




    $begingroup$
    math.stackexchange.com/questions/123524/…
    $endgroup$
    – lab bhattacharjee
    Mar 14 at 4:35






  • 1




    $begingroup$
    I assume that you also want that $x neq y$, else the gcd of the $2$ values will be the values them.
    $endgroup$
    – John Omielan
    Mar 14 at 4:37







3




3




$begingroup$
math.stackexchange.com/questions/123524/…
$endgroup$
– lab bhattacharjee
Mar 14 at 4:35




$begingroup$
math.stackexchange.com/questions/123524/…
$endgroup$
– lab bhattacharjee
Mar 14 at 4:35




1




1




$begingroup$
I assume that you also want that $x neq y$, else the gcd of the $2$ values will be the values them.
$endgroup$
– John Omielan
Mar 14 at 4:37




$begingroup$
I assume that you also want that $x neq y$, else the gcd of the $2$ values will be the values them.
$endgroup$
– John Omielan
Mar 14 at 4:37










1 Answer
1






active

oldest

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2












$begingroup$

Let $x<y.$ Let $n=gcd(1+z^2^x,1+2^2^y).$ Then $z^2^xequiv -1 equiv z^2^y pmod n.$ Now $y-x-1$ is a non-negative integer, so modulo $n$ we have $$-1equiv z^2^yequiv (,(z^2^x)^2^y-x-1,)^2equiv$$ $$equiv (,(-1)^2^y-x-1,)^2equiv$$ $$equiv (,pm 1,)^2 equiv 1.$$



So $-1equiv 1 pmod n.$ And since $dequiv e pmod n iff n|(e-d)$ (for any $d,e$), we have $n|(1-(-1))=2,$ so $nle 2.$



We have used the fact that for any $a,b, nin Bbb Z$ with $nne 0,$ and any $cin Bbb N,$ if $aequiv b pmod n$ then $a^cequiv b^c pmod n.$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    This is also valid, verbatim, if $ 0=x<y.$
    $endgroup$
    – DanielWainfleet
    Mar 14 at 6:48










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1 Answer
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1 Answer
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2












$begingroup$

Let $x<y.$ Let $n=gcd(1+z^2^x,1+2^2^y).$ Then $z^2^xequiv -1 equiv z^2^y pmod n.$ Now $y-x-1$ is a non-negative integer, so modulo $n$ we have $$-1equiv z^2^yequiv (,(z^2^x)^2^y-x-1,)^2equiv$$ $$equiv (,(-1)^2^y-x-1,)^2equiv$$ $$equiv (,pm 1,)^2 equiv 1.$$



So $-1equiv 1 pmod n.$ And since $dequiv e pmod n iff n|(e-d)$ (for any $d,e$), we have $n|(1-(-1))=2,$ so $nle 2.$



We have used the fact that for any $a,b, nin Bbb Z$ with $nne 0,$ and any $cin Bbb N,$ if $aequiv b pmod n$ then $a^cequiv b^c pmod n.$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    This is also valid, verbatim, if $ 0=x<y.$
    $endgroup$
    – DanielWainfleet
    Mar 14 at 6:48















2












$begingroup$

Let $x<y.$ Let $n=gcd(1+z^2^x,1+2^2^y).$ Then $z^2^xequiv -1 equiv z^2^y pmod n.$ Now $y-x-1$ is a non-negative integer, so modulo $n$ we have $$-1equiv z^2^yequiv (,(z^2^x)^2^y-x-1,)^2equiv$$ $$equiv (,(-1)^2^y-x-1,)^2equiv$$ $$equiv (,pm 1,)^2 equiv 1.$$



So $-1equiv 1 pmod n.$ And since $dequiv e pmod n iff n|(e-d)$ (for any $d,e$), we have $n|(1-(-1))=2,$ so $nle 2.$



We have used the fact that for any $a,b, nin Bbb Z$ with $nne 0,$ and any $cin Bbb N,$ if $aequiv b pmod n$ then $a^cequiv b^c pmod n.$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    This is also valid, verbatim, if $ 0=x<y.$
    $endgroup$
    – DanielWainfleet
    Mar 14 at 6:48













2












2








2





$begingroup$

Let $x<y.$ Let $n=gcd(1+z^2^x,1+2^2^y).$ Then $z^2^xequiv -1 equiv z^2^y pmod n.$ Now $y-x-1$ is a non-negative integer, so modulo $n$ we have $$-1equiv z^2^yequiv (,(z^2^x)^2^y-x-1,)^2equiv$$ $$equiv (,(-1)^2^y-x-1,)^2equiv$$ $$equiv (,pm 1,)^2 equiv 1.$$



So $-1equiv 1 pmod n.$ And since $dequiv e pmod n iff n|(e-d)$ (for any $d,e$), we have $n|(1-(-1))=2,$ so $nle 2.$



We have used the fact that for any $a,b, nin Bbb Z$ with $nne 0,$ and any $cin Bbb N,$ if $aequiv b pmod n$ then $a^cequiv b^c pmod n.$






share|cite|improve this answer











$endgroup$



Let $x<y.$ Let $n=gcd(1+z^2^x,1+2^2^y).$ Then $z^2^xequiv -1 equiv z^2^y pmod n.$ Now $y-x-1$ is a non-negative integer, so modulo $n$ we have $$-1equiv z^2^yequiv (,(z^2^x)^2^y-x-1,)^2equiv$$ $$equiv (,(-1)^2^y-x-1,)^2equiv$$ $$equiv (,pm 1,)^2 equiv 1.$$



So $-1equiv 1 pmod n.$ And since $dequiv e pmod n iff n|(e-d)$ (for any $d,e$), we have $n|(1-(-1))=2,$ so $nle 2.$



We have used the fact that for any $a,b, nin Bbb Z$ with $nne 0,$ and any $cin Bbb N,$ if $aequiv b pmod n$ then $a^cequiv b^c pmod n.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 14 at 6:54

























answered Mar 14 at 6:46









DanielWainfleetDanielWainfleet

35.6k31648




35.6k31648











  • $begingroup$
    This is also valid, verbatim, if $ 0=x<y.$
    $endgroup$
    – DanielWainfleet
    Mar 14 at 6:48
















  • $begingroup$
    This is also valid, verbatim, if $ 0=x<y.$
    $endgroup$
    – DanielWainfleet
    Mar 14 at 6:48















$begingroup$
This is also valid, verbatim, if $ 0=x<y.$
$endgroup$
– DanielWainfleet
Mar 14 at 6:48




$begingroup$
This is also valid, verbatim, if $ 0=x<y.$
$endgroup$
– DanielWainfleet
Mar 14 at 6:48

















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