Equation solving?












0














I've got this system of equations.
(a,b,c) ∈ ℤ



$a*b+1 = c$



$a^2 + b^2 +1 = 2c$



$2a + b = c$



I tried to substitute a little bit:



$a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$



Ultimately:



$a(a-2-b) + b(b-1) = 0$



Now I'm not sure how to substitute b into a in order to go on.



Maybe I didn't even start well.
Can someone help?



Thx






systems-of-equations quadratics







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    0














    I've got this system of equations.
    (a,b,c) ∈ ℤ



    $a*b+1 = c$



    $a^2 + b^2 +1 = 2c$



    $2a + b = c$



    I tried to substitute a little bit:



    $a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$



    Ultimately:



    $a(a-2-b) + b(b-1) = 0$



    Now I'm not sure how to substitute b into a in order to go on.



    Maybe I didn't even start well.
    Can someone help?



    Thx






    systems-of-equations quadratics







    share|cite|improve this question



























      0












      0








      0







      I've got this system of equations.
      (a,b,c) ∈ ℤ



      $a*b+1 = c$



      $a^2 + b^2 +1 = 2c$



      $2a + b = c$



      I tried to substitute a little bit:



      $a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$



      Ultimately:



      $a(a-2-b) + b(b-1) = 0$



      Now I'm not sure how to substitute b into a in order to go on.



      Maybe I didn't even start well.
      Can someone help?



      Thx






      systems-of-equations quadratics







      share|cite|improve this question















      I've got this system of equations.
      (a,b,c) ∈ ℤ



      $a*b+1 = c$



      $a^2 + b^2 +1 = 2c$



      $2a + b = c$



      I tried to substitute a little bit:



      $a^2 + b^2 + 1 -(2a+b) - (a*b+1) = 0$



      Ultimately:



      $a(a-2-b) + b(b-1) = 0$



      Now I'm not sure how to substitute b into a in order to go on.



      Maybe I didn't even start well.
      Can someone help?



      Thx






      systems-of-equations quadratics




      systems-of-equations quadratics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 23 '18 at 12:00









      Harry Peter

      5,46111439




      5,46111439










      asked Nov 15 '18 at 11:30









      calculatormathematicalcalculatormathematical

      3811




      3811






















          3 Answers
          3






          active

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          2














          You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.






          share|cite|improve this answer





























            0














            Eliminating $c$ from the first two equations,



            $$(a-b)^2=1$$ or $$a=bpm1.$$



            Then eliminating $a$ and $c$ from the first and third, we get two quadratic,



            $$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$



            $$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$






            share|cite|improve this answer





















            • Ok but what if (a,b,c) ∈ ℤ
              – calculatormathematical
              Nov 15 '18 at 11:54










            • @calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
              – Yves Daoust
              Nov 15 '18 at 14:44



















            0














            Here is a possible way to solve your problem.



            From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
            Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
            I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.



            Hope it works.






            share|cite|improve this answer





















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.






              share|cite|improve this answer


























                2














                You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.






                  share|cite|improve this answer












                  You can obtain $|a-b|=1$ by substracting the first equation (where you multiply by 2 both left and right-hand side) from the second equation, so you get $a^2+b^2-2ab -1 = 0$. From this, you can write for instance $b$ and $c$ in function of $a$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 15 '18 at 11:41









                  LacramioaraLacramioara

                  32126




                  32126























                      0














                      Eliminating $c$ from the first two equations,



                      $$(a-b)^2=1$$ or $$a=bpm1.$$



                      Then eliminating $a$ and $c$ from the first and third, we get two quadratic,



                      $$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$



                      $$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$






                      share|cite|improve this answer





















                      • Ok but what if (a,b,c) ∈ ℤ
                        – calculatormathematical
                        Nov 15 '18 at 11:54










                      • @calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
                        – Yves Daoust
                        Nov 15 '18 at 14:44
















                      0














                      Eliminating $c$ from the first two equations,



                      $$(a-b)^2=1$$ or $$a=bpm1.$$



                      Then eliminating $a$ and $c$ from the first and third, we get two quadratic,



                      $$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$



                      $$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$






                      share|cite|improve this answer





















                      • Ok but what if (a,b,c) ∈ ℤ
                        – calculatormathematical
                        Nov 15 '18 at 11:54










                      • @calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
                        – Yves Daoust
                        Nov 15 '18 at 14:44














                      0












                      0








                      0






                      Eliminating $c$ from the first two equations,



                      $$(a-b)^2=1$$ or $$a=bpm1.$$



                      Then eliminating $a$ and $c$ from the first and third, we get two quadratic,



                      $$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$



                      $$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$






                      share|cite|improve this answer












                      Eliminating $c$ from the first two equations,



                      $$(a-b)^2=1$$ or $$a=bpm1.$$



                      Then eliminating $a$ and $c$ from the first and third, we get two quadratic,



                      $$(b+1)b+1=2(b+1)+b\to b=1pmsqrt2,a=2pmsqrt2,c=5pm3sqrt2,$$



                      $$(b-1)b+1=2(b-1)+b\to b=1,3, a=0,2, c=1,7.$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 15 '18 at 11:38









                      Yves DaoustYves Daoust

                      124k671222




                      124k671222












                      • Ok but what if (a,b,c) ∈ ℤ
                        – calculatormathematical
                        Nov 15 '18 at 11:54










                      • @calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
                        – Yves Daoust
                        Nov 15 '18 at 14:44


















                      • Ok but what if (a,b,c) ∈ ℤ
                        – calculatormathematical
                        Nov 15 '18 at 11:54










                      • @calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
                        – Yves Daoust
                        Nov 15 '18 at 14:44
















                      Ok but what if (a,b,c) ∈ ℤ
                      – calculatormathematical
                      Nov 15 '18 at 11:54




                      Ok but what if (a,b,c) ∈ ℤ
                      – calculatormathematical
                      Nov 15 '18 at 11:54












                      @calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
                      – Yves Daoust
                      Nov 15 '18 at 14:44




                      @calculatormathematical: this constraint was added later. Can you tell an irrational from an integer ?
                      – Yves Daoust
                      Nov 15 '18 at 14:44











                      0














                      Here is a possible way to solve your problem.



                      From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
                      Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
                      I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.



                      Hope it works.






                      share|cite|improve this answer


























                        0














                        Here is a possible way to solve your problem.



                        From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
                        Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
                        I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.



                        Hope it works.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Here is a possible way to solve your problem.



                          From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
                          Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
                          I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.



                          Hope it works.






                          share|cite|improve this answer












                          Here is a possible way to solve your problem.



                          From the first and second equation, we have, $$a^2+b^2+1=2ab+2\or,(a-b)^2=1\or,a-b=pm1$$
                          Now, using 3rd equation we get, $$a=frac{1}{3}(cpm 1)$$ So, $$b=amp1=frac{1}{3}(cmp 2)$$
                          I have skipped the steps hope you find them. If not then let me know. Further, here I assumed that $a,b,c$ are real numbers and the meaning of $mp=-(pm)$.



                          Hope it works.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 15 '18 at 11:40









                          Sujit BhattacharyyaSujit Bhattacharyya

                          945318




                          945318






























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