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The law of quadratic reciprocity 'or' statement


quadratic reciprocityShow that the statement in the Law of Quadratic Reciprocity can be written (as Gauss did as)Equivalent statement to Quadratic reciprocityDifferent formulations of the Law of Quadratic ReciprocityQuadratic ReciprocityAbout a Proof of Quadratic Reciprocity LawSecond supplemental law of Eisenstein ReciprocityUsing quadratic reciprocity to motivate higher reciprocity laws?How do I use the Law of Quadratic Reciprocity to solve a congruence?Kronecker symbol and the connection to quadratic reciprocity













0












$begingroup$


I have just seen written that $(frac17101)$ = $(frac10117)$. But doesn't the law of quadratic reciprocity state that $(fracpq)$ = $(fracqp)$ if $pequiv1 mod 4$ OR $qequiv1mod4$?










share|cite|improve this question









$endgroup$







  • 7




    $begingroup$
    In mathematics, "or" is almost always inclusive, so we allow for both conditions to hold.
    $endgroup$
    – Wojowu
    yesterday
















0












$begingroup$


I have just seen written that $(frac17101)$ = $(frac10117)$. But doesn't the law of quadratic reciprocity state that $(fracpq)$ = $(fracqp)$ if $pequiv1 mod 4$ OR $qequiv1mod4$?










share|cite|improve this question









$endgroup$







  • 7




    $begingroup$
    In mathematics, "or" is almost always inclusive, so we allow for both conditions to hold.
    $endgroup$
    – Wojowu
    yesterday














0












0








0





$begingroup$


I have just seen written that $(frac17101)$ = $(frac10117)$. But doesn't the law of quadratic reciprocity state that $(fracpq)$ = $(fracqp)$ if $pequiv1 mod 4$ OR $qequiv1mod4$?










share|cite|improve this question









$endgroup$




I have just seen written that $(frac17101)$ = $(frac10117)$. But doesn't the law of quadratic reciprocity state that $(fracpq)$ = $(fracqp)$ if $pequiv1 mod 4$ OR $qequiv1mod4$?







number-theory modular-arithmetic legendre-symbol






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









JlatmerJlatmer

64




64







  • 7




    $begingroup$
    In mathematics, "or" is almost always inclusive, so we allow for both conditions to hold.
    $endgroup$
    – Wojowu
    yesterday













  • 7




    $begingroup$
    In mathematics, "or" is almost always inclusive, so we allow for both conditions to hold.
    $endgroup$
    – Wojowu
    yesterday








7




7




$begingroup$
In mathematics, "or" is almost always inclusive, so we allow for both conditions to hold.
$endgroup$
– Wojowu
yesterday





$begingroup$
In mathematics, "or" is almost always inclusive, so we allow for both conditions to hold.
$endgroup$
– Wojowu
yesterday











0






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