Let a, b, c, d be any four distinct integers such that $a>b>c>d>1$. Show that if $ad=bc$, then...












0












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Let $a, b, c, d$ be any four distinct integers such that $a>b>c>d>1$.



Show that if $ad=bc$, then $a^2+b^2+c^2+d^2$ is composite.







elementary-number-theory divisibility sums-of-squares







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closed as off-topic by user21820, Holo, Did, Xander Henderson, user98602 Feb 2 at 15:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, Xander Henderson, Community

If this question can be reworded to fit the rules in the help center, please edit the question.





















    0












    $begingroup$



    Let $a, b, c, d$ be any four distinct integers such that $a>b>c>d>1$.



    Show that if $ad=bc$, then $a^2+b^2+c^2+d^2$ is composite.







    elementary-number-theory divisibility sums-of-squares







    share|cite|improve this question











    $endgroup$



    closed as off-topic by user21820, Holo, Did, Xander Henderson, user98602 Feb 2 at 15:56


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, Xander Henderson, Community

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0





      $begingroup$



      Let $a, b, c, d$ be any four distinct integers such that $a>b>c>d>1$.



      Show that if $ad=bc$, then $a^2+b^2+c^2+d^2$ is composite.







      elementary-number-theory divisibility sums-of-squares







      share|cite|improve this question











      $endgroup$





      Let $a, b, c, d$ be any four distinct integers such that $a>b>c>d>1$.



      Show that if $ad=bc$, then $a^2+b^2+c^2+d^2$ is composite.







      elementary-number-theory divisibility sums-of-squares




      elementary-number-theory divisibility sums-of-squares






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 27 at 10:56









      Martin Sleziak

      44.9k10122276




      44.9k10122276










      asked Jan 27 at 8:51









      Anson ChanAnson Chan

      192




      192




      closed as off-topic by user21820, Holo, Did, Xander Henderson, user98602 Feb 2 at 15:56


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, Xander Henderson, Community

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by user21820, Holo, Did, Xander Henderson, user98602 Feb 2 at 15:56


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Holo, Did, Xander Henderson, Community

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          Write $d={bcover a}$ and suppose $a^2+b^2+c^2+d^2=p$ is prime.



          So $$ a^4+a^2(b^2+c^2)+b^2c^2 = pa^2$$



          but now $$(a^2+b^2)(a^2+c^2)= pa^2$$ which implies $$pmid a^2+b^2;;;{rm or};;;pmid a^2+c^2$$



          But then we get $a^2+b^2+c^2+d^2 = pleq a^2+b^2$ or $a^2+b^2+c^2+d^2 = pleq a^2+c^2$ which is impossible. So $p$ must be composite.



          Note that we did not need arrangment of numbers.






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            It is easy:
            $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{a}{b}right)^2 + 1right) b^2 + left(left(frac{c}{d}right)^2 + 1right) d^2$$



            As $frac{a}{b} = frac{c}{d}$:
            $$A = left(left(frac{a}{b}right)^2 + 1right) left(b^2 + d^2right)$$



            Also:



            $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{b}{a}right)^2 + 1right) a^2 + left(left(frac{d}{c}right)^2 + 1right) c^2 = left(left(frac{b}{a}right)^2 + 1right) left(a^2 + c^2right) $$



            Hence:



            $$frac{a^2 + b^2}{a^2} left(a^2 + c^2right) = frac{a^2 + b^2}{b^2} left(b^2 + d^2right)Rightarrow frac{a^2 + c^2}{a^2} = frac{b^2 + d^2}{b^2}Rightarrow a^2 + c^2 = frac{(b^2 + d^2)a^2}{b^2}Rightarrow \ b|d text{ or } b|a text{ or } b = pq text{ such that } p | a, q|d$$.



            As $b > d$, $b | a$ or the thrid case. For $b | a$, $frac{a}{b}$ is integer, and $A$ is composite. For the third case we have:



            $$A = left(frac{a^2+ b^2}{p^2}right)left(frac{b^2+d^2}{q^2}right)$$



            As $p|a$ and $q|d$ we can say $A$ is composite in this case.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How do you know (a/b) is an integer?
              $endgroup$
              – Erik Parkinson
              Jan 27 at 8:56












            • $begingroup$
              I have the same question as Erik.
              $endgroup$
              – Maria Mazur
              Jan 27 at 9:03










            • $begingroup$
              @ErikParkinson it's updated.
              $endgroup$
              – OmG
              Jan 27 at 9:08












            • $begingroup$
              @greedoid it's updated.
              $endgroup$
              – OmG
              Jan 27 at 9:08










            • $begingroup$
              Awesome, looks great!
              $endgroup$
              – Erik Parkinson
              Jan 27 at 9:10



















            0












            $begingroup$

            Since $amid bc$, we know that $left.frac{a}{(a,b)}middle|frac{b}{(a,b)}cright.$. Since $frac{a}{(a,b)}$ and $frac{b}{(a,b)}$ are relatively prime, we have $left.frac{a}{(a,b)}middle|,cright.$. Therefore,
            $$
            bbox[5px,border:2px solid #C0A000]{a^n+b^n+c^n+d^n=left(frac{b^n}{(a,b)^n}+frac{a^n}{(a,b)^n}right)left((a,b)^n+frac{c^n(a,b)^n}{a^n}right)}tag1
            $$






            share|cite|improve this answer











            $endgroup$




















              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              Write $d={bcover a}$ and suppose $a^2+b^2+c^2+d^2=p$ is prime.



              So $$ a^4+a^2(b^2+c^2)+b^2c^2 = pa^2$$



              but now $$(a^2+b^2)(a^2+c^2)= pa^2$$ which implies $$pmid a^2+b^2;;;{rm or};;;pmid a^2+c^2$$



              But then we get $a^2+b^2+c^2+d^2 = pleq a^2+b^2$ or $a^2+b^2+c^2+d^2 = pleq a^2+c^2$ which is impossible. So $p$ must be composite.



              Note that we did not need arrangment of numbers.






              share|cite|improve this answer











              $endgroup$


















                5












                $begingroup$

                Write $d={bcover a}$ and suppose $a^2+b^2+c^2+d^2=p$ is prime.



                So $$ a^4+a^2(b^2+c^2)+b^2c^2 = pa^2$$



                but now $$(a^2+b^2)(a^2+c^2)= pa^2$$ which implies $$pmid a^2+b^2;;;{rm or};;;pmid a^2+c^2$$



                But then we get $a^2+b^2+c^2+d^2 = pleq a^2+b^2$ or $a^2+b^2+c^2+d^2 = pleq a^2+c^2$ which is impossible. So $p$ must be composite.



                Note that we did not need arrangment of numbers.






                share|cite|improve this answer











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Write $d={bcover a}$ and suppose $a^2+b^2+c^2+d^2=p$ is prime.



                  So $$ a^4+a^2(b^2+c^2)+b^2c^2 = pa^2$$



                  but now $$(a^2+b^2)(a^2+c^2)= pa^2$$ which implies $$pmid a^2+b^2;;;{rm or};;;pmid a^2+c^2$$



                  But then we get $a^2+b^2+c^2+d^2 = pleq a^2+b^2$ or $a^2+b^2+c^2+d^2 = pleq a^2+c^2$ which is impossible. So $p$ must be composite.



                  Note that we did not need arrangment of numbers.






                  share|cite|improve this answer











                  $endgroup$



                  Write $d={bcover a}$ and suppose $a^2+b^2+c^2+d^2=p$ is prime.



                  So $$ a^4+a^2(b^2+c^2)+b^2c^2 = pa^2$$



                  but now $$(a^2+b^2)(a^2+c^2)= pa^2$$ which implies $$pmid a^2+b^2;;;{rm or};;;pmid a^2+c^2$$



                  But then we get $a^2+b^2+c^2+d^2 = pleq a^2+b^2$ or $a^2+b^2+c^2+d^2 = pleq a^2+c^2$ which is impossible. So $p$ must be composite.



                  Note that we did not need arrangment of numbers.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 27 at 13:08

























                  answered Jan 27 at 9:02









                  Maria MazurMaria Mazur

                  48.3k1260121




                  48.3k1260121























                      3












                      $begingroup$

                      It is easy:
                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{a}{b}right)^2 + 1right) b^2 + left(left(frac{c}{d}right)^2 + 1right) d^2$$



                      As $frac{a}{b} = frac{c}{d}$:
                      $$A = left(left(frac{a}{b}right)^2 + 1right) left(b^2 + d^2right)$$



                      Also:



                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{b}{a}right)^2 + 1right) a^2 + left(left(frac{d}{c}right)^2 + 1right) c^2 = left(left(frac{b}{a}right)^2 + 1right) left(a^2 + c^2right) $$



                      Hence:



                      $$frac{a^2 + b^2}{a^2} left(a^2 + c^2right) = frac{a^2 + b^2}{b^2} left(b^2 + d^2right)Rightarrow frac{a^2 + c^2}{a^2} = frac{b^2 + d^2}{b^2}Rightarrow a^2 + c^2 = frac{(b^2 + d^2)a^2}{b^2}Rightarrow \ b|d text{ or } b|a text{ or } b = pq text{ such that } p | a, q|d$$.



                      As $b > d$, $b | a$ or the thrid case. For $b | a$, $frac{a}{b}$ is integer, and $A$ is composite. For the third case we have:



                      $$A = left(frac{a^2+ b^2}{p^2}right)left(frac{b^2+d^2}{q^2}right)$$



                      As $p|a$ and $q|d$ we can say $A$ is composite in this case.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        How do you know (a/b) is an integer?
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 8:56












                      • $begingroup$
                        I have the same question as Erik.
                        $endgroup$
                        – Maria Mazur
                        Jan 27 at 9:03










                      • $begingroup$
                        @ErikParkinson it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08












                      • $begingroup$
                        @greedoid it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08










                      • $begingroup$
                        Awesome, looks great!
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 9:10
















                      3












                      $begingroup$

                      It is easy:
                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{a}{b}right)^2 + 1right) b^2 + left(left(frac{c}{d}right)^2 + 1right) d^2$$



                      As $frac{a}{b} = frac{c}{d}$:
                      $$A = left(left(frac{a}{b}right)^2 + 1right) left(b^2 + d^2right)$$



                      Also:



                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{b}{a}right)^2 + 1right) a^2 + left(left(frac{d}{c}right)^2 + 1right) c^2 = left(left(frac{b}{a}right)^2 + 1right) left(a^2 + c^2right) $$



                      Hence:



                      $$frac{a^2 + b^2}{a^2} left(a^2 + c^2right) = frac{a^2 + b^2}{b^2} left(b^2 + d^2right)Rightarrow frac{a^2 + c^2}{a^2} = frac{b^2 + d^2}{b^2}Rightarrow a^2 + c^2 = frac{(b^2 + d^2)a^2}{b^2}Rightarrow \ b|d text{ or } b|a text{ or } b = pq text{ such that } p | a, q|d$$.



                      As $b > d$, $b | a$ or the thrid case. For $b | a$, $frac{a}{b}$ is integer, and $A$ is composite. For the third case we have:



                      $$A = left(frac{a^2+ b^2}{p^2}right)left(frac{b^2+d^2}{q^2}right)$$



                      As $p|a$ and $q|d$ we can say $A$ is composite in this case.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        How do you know (a/b) is an integer?
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 8:56












                      • $begingroup$
                        I have the same question as Erik.
                        $endgroup$
                        – Maria Mazur
                        Jan 27 at 9:03










                      • $begingroup$
                        @ErikParkinson it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08












                      • $begingroup$
                        @greedoid it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08










                      • $begingroup$
                        Awesome, looks great!
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 9:10














                      3












                      3








                      3





                      $begingroup$

                      It is easy:
                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{a}{b}right)^2 + 1right) b^2 + left(left(frac{c}{d}right)^2 + 1right) d^2$$



                      As $frac{a}{b} = frac{c}{d}$:
                      $$A = left(left(frac{a}{b}right)^2 + 1right) left(b^2 + d^2right)$$



                      Also:



                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{b}{a}right)^2 + 1right) a^2 + left(left(frac{d}{c}right)^2 + 1right) c^2 = left(left(frac{b}{a}right)^2 + 1right) left(a^2 + c^2right) $$



                      Hence:



                      $$frac{a^2 + b^2}{a^2} left(a^2 + c^2right) = frac{a^2 + b^2}{b^2} left(b^2 + d^2right)Rightarrow frac{a^2 + c^2}{a^2} = frac{b^2 + d^2}{b^2}Rightarrow a^2 + c^2 = frac{(b^2 + d^2)a^2}{b^2}Rightarrow \ b|d text{ or } b|a text{ or } b = pq text{ such that } p | a, q|d$$.



                      As $b > d$, $b | a$ or the thrid case. For $b | a$, $frac{a}{b}$ is integer, and $A$ is composite. For the third case we have:



                      $$A = left(frac{a^2+ b^2}{p^2}right)left(frac{b^2+d^2}{q^2}right)$$



                      As $p|a$ and $q|d$ we can say $A$ is composite in this case.






                      share|cite|improve this answer











                      $endgroup$



                      It is easy:
                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{a}{b}right)^2 + 1right) b^2 + left(left(frac{c}{d}right)^2 + 1right) d^2$$



                      As $frac{a}{b} = frac{c}{d}$:
                      $$A = left(left(frac{a}{b}right)^2 + 1right) left(b^2 + d^2right)$$



                      Also:



                      $$A = a^2 + b^2 + c^2 + d^2 = left(left(frac{b}{a}right)^2 + 1right) a^2 + left(left(frac{d}{c}right)^2 + 1right) c^2 = left(left(frac{b}{a}right)^2 + 1right) left(a^2 + c^2right) $$



                      Hence:



                      $$frac{a^2 + b^2}{a^2} left(a^2 + c^2right) = frac{a^2 + b^2}{b^2} left(b^2 + d^2right)Rightarrow frac{a^2 + c^2}{a^2} = frac{b^2 + d^2}{b^2}Rightarrow a^2 + c^2 = frac{(b^2 + d^2)a^2}{b^2}Rightarrow \ b|d text{ or } b|a text{ or } b = pq text{ such that } p | a, q|d$$.



                      As $b > d$, $b | a$ or the thrid case. For $b | a$, $frac{a}{b}$ is integer, and $A$ is composite. For the third case we have:



                      $$A = left(frac{a^2+ b^2}{p^2}right)left(frac{b^2+d^2}{q^2}right)$$



                      As $p|a$ and $q|d$ we can say $A$ is composite in this case.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 28 at 13:01

























                      answered Jan 27 at 8:55









                      OmGOmG

                      2,512824




                      2,512824












                      • $begingroup$
                        How do you know (a/b) is an integer?
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 8:56












                      • $begingroup$
                        I have the same question as Erik.
                        $endgroup$
                        – Maria Mazur
                        Jan 27 at 9:03










                      • $begingroup$
                        @ErikParkinson it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08












                      • $begingroup$
                        @greedoid it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08










                      • $begingroup$
                        Awesome, looks great!
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 9:10


















                      • $begingroup$
                        How do you know (a/b) is an integer?
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 8:56












                      • $begingroup$
                        I have the same question as Erik.
                        $endgroup$
                        – Maria Mazur
                        Jan 27 at 9:03










                      • $begingroup$
                        @ErikParkinson it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08












                      • $begingroup$
                        @greedoid it's updated.
                        $endgroup$
                        – OmG
                        Jan 27 at 9:08










                      • $begingroup$
                        Awesome, looks great!
                        $endgroup$
                        – Erik Parkinson
                        Jan 27 at 9:10
















                      $begingroup$
                      How do you know (a/b) is an integer?
                      $endgroup$
                      – Erik Parkinson
                      Jan 27 at 8:56






                      $begingroup$
                      How do you know (a/b) is an integer?
                      $endgroup$
                      – Erik Parkinson
                      Jan 27 at 8:56














                      $begingroup$
                      I have the same question as Erik.
                      $endgroup$
                      – Maria Mazur
                      Jan 27 at 9:03




                      $begingroup$
                      I have the same question as Erik.
                      $endgroup$
                      – Maria Mazur
                      Jan 27 at 9:03












                      $begingroup$
                      @ErikParkinson it's updated.
                      $endgroup$
                      – OmG
                      Jan 27 at 9:08






                      $begingroup$
                      @ErikParkinson it's updated.
                      $endgroup$
                      – OmG
                      Jan 27 at 9:08














                      $begingroup$
                      @greedoid it's updated.
                      $endgroup$
                      – OmG
                      Jan 27 at 9:08




                      $begingroup$
                      @greedoid it's updated.
                      $endgroup$
                      – OmG
                      Jan 27 at 9:08












                      $begingroup$
                      Awesome, looks great!
                      $endgroup$
                      – Erik Parkinson
                      Jan 27 at 9:10




                      $begingroup$
                      Awesome, looks great!
                      $endgroup$
                      – Erik Parkinson
                      Jan 27 at 9:10











                      0












                      $begingroup$

                      Since $amid bc$, we know that $left.frac{a}{(a,b)}middle|frac{b}{(a,b)}cright.$. Since $frac{a}{(a,b)}$ and $frac{b}{(a,b)}$ are relatively prime, we have $left.frac{a}{(a,b)}middle|,cright.$. Therefore,
                      $$
                      bbox[5px,border:2px solid #C0A000]{a^n+b^n+c^n+d^n=left(frac{b^n}{(a,b)^n}+frac{a^n}{(a,b)^n}right)left((a,b)^n+frac{c^n(a,b)^n}{a^n}right)}tag1
                      $$






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Since $amid bc$, we know that $left.frac{a}{(a,b)}middle|frac{b}{(a,b)}cright.$. Since $frac{a}{(a,b)}$ and $frac{b}{(a,b)}$ are relatively prime, we have $left.frac{a}{(a,b)}middle|,cright.$. Therefore,
                        $$
                        bbox[5px,border:2px solid #C0A000]{a^n+b^n+c^n+d^n=left(frac{b^n}{(a,b)^n}+frac{a^n}{(a,b)^n}right)left((a,b)^n+frac{c^n(a,b)^n}{a^n}right)}tag1
                        $$






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Since $amid bc$, we know that $left.frac{a}{(a,b)}middle|frac{b}{(a,b)}cright.$. Since $frac{a}{(a,b)}$ and $frac{b}{(a,b)}$ are relatively prime, we have $left.frac{a}{(a,b)}middle|,cright.$. Therefore,
                          $$
                          bbox[5px,border:2px solid #C0A000]{a^n+b^n+c^n+d^n=left(frac{b^n}{(a,b)^n}+frac{a^n}{(a,b)^n}right)left((a,b)^n+frac{c^n(a,b)^n}{a^n}right)}tag1
                          $$






                          share|cite|improve this answer











                          $endgroup$



                          Since $amid bc$, we know that $left.frac{a}{(a,b)}middle|frac{b}{(a,b)}cright.$. Since $frac{a}{(a,b)}$ and $frac{b}{(a,b)}$ are relatively prime, we have $left.frac{a}{(a,b)}middle|,cright.$. Therefore,
                          $$
                          bbox[5px,border:2px solid #C0A000]{a^n+b^n+c^n+d^n=left(frac{b^n}{(a,b)^n}+frac{a^n}{(a,b)^n}right)left((a,b)^n+frac{c^n(a,b)^n}{a^n}right)}tag1
                          $$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 27 at 15:18

























                          answered Jan 27 at 14:59









                          robjohn♦robjohn

                          270k27311639




                          270k27311639















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                              MongoDB - Not Authorized To Execute Command

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                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
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                              How to fix TextFormField cause rebuild widget in Flutter

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                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
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                              in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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                              1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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