if $a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$

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If $a+b$ is an positive integer and $age b$ and
$a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.




I tried a lot to solve it, i came to a step after which i was not able to proceed forward, the step was $(a+b)^3 - 41cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$
Please help me to solve this.




algebra-precalculus number-theory




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  • Please use MathJax instead of relying on others to format your question.
    – Toby Mak
    2 days ago










  • Hmm, perhaps you can get some progress out of noticing that $6ab$ is a multiple of $a+b$.
    – Henning Makholm
    2 days ago










  • $ab$ may be non integer.
    – Takahiro Waki
    6 hours ago














up vote
2
down vote

favorite
2













If $a+b$ is an positive integer and $age b$ and
$a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.




I tried a lot to solve it, i came to a step after which i was not able to proceed forward, the step was $(a+b)^3 - 41cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$
Please help me to solve this.




algebra-precalculus number-theory




share|cite|improve this question





















  • Please use MathJax instead of relying on others to format your question.
    – Toby Mak
    2 days ago










  • Hmm, perhaps you can get some progress out of noticing that $6ab$ is a multiple of $a+b$.
    – Henning Makholm
    2 days ago










  • $ab$ may be non integer.
    – Takahiro Waki
    6 hours ago












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2






If $a+b$ is an positive integer and $age b$ and
$a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.




I tried a lot to solve it, i came to a step after which i was not able to proceed forward, the step was $(a+b)^3 - 41cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$
Please help me to solve this.




algebra-precalculus number-theory




share|cite|improve this question














If $a+b$ is an positive integer and $age b$ and
$a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.




I tried a lot to solve it, i came to a step after which i was not able to proceed forward, the step was $(a+b)^3 - 41cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$
Please help me to solve this.





algebra-precalculus number-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Henning Makholm

225k16289516




225k16289516









asked 2 days ago









Smit Patel

394




394











  • Please use MathJax instead of relying on others to format your question.
    – Toby Mak
    2 days ago










  • Hmm, perhaps you can get some progress out of noticing that $6ab$ is a multiple of $a+b$.
    – Henning Makholm
    2 days ago










  • $ab$ may be non integer.
    – Takahiro Waki
    6 hours ago
















  • Please use MathJax instead of relying on others to format your question.
    – Toby Mak
    2 days ago










  • Hmm, perhaps you can get some progress out of noticing that $6ab$ is a multiple of $a+b$.
    – Henning Makholm
    2 days ago










  • $ab$ may be non integer.
    – Takahiro Waki
    6 hours ago















Please use MathJax instead of relying on others to format your question.
– Toby Mak
2 days ago




Please use MathJax instead of relying on others to format your question.
– Toby Mak
2 days ago












Hmm, perhaps you can get some progress out of noticing that $6ab$ is a multiple of $a+b$.
– Henning Makholm
2 days ago




Hmm, perhaps you can get some progress out of noticing that $6ab$ is a multiple of $a+b$.
– Henning Makholm
2 days ago












$ab$ may be non integer.
– Takahiro Waki
6 hours ago




$ab$ may be non integer.
– Takahiro Waki
6 hours ago










2 Answers
2






active

oldest

votes

















up vote
1
down vote













Note: I solve it base on that a,b are positive numbers(not integers)



suppose $s=a+b$



Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$



Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$



$Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$



$0<s<=2794$



for one solution, $0<a<s$



(1) $0<3s^2+6s-sqrtDelta<2s(6+3s)$



L:$3s^2+6s>sqrtDelta$



$(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$



$4(6+3s)(s^3-697s^2)>0$



$0<s<697$



R:$-3s^2-6s<sqrtDelta$ It is always true



(2) $0<3s^2+6s+sqrtDelta<2s(6+3s)$



L: always true



R: $3s^2+6s>sqrtDelta$



It is true for $0<s<697$



Overall $a+b=s=1,2,...695$ or $696$



$sum=242556$






share|cite|improve this answer























  • @RossMillikan $a$ and $b$ can be noninteger
    – Mira from Earth
    2 days ago










  • I now see that.
    – Ross Millikan
    2 days ago

















up vote
0
down vote













Here is a start, not a full answer.



Let $s=a+b$ and $p=ab$. Then $3 p (s + 2) = (s - 697) s^2$ as WA tells us.



Then $s+2$ divides $(s - 697) s^2$ and so $s+2$ divides $2796 = 2^2 cdot 3 cdot 233$.






share|cite|improve this answer























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    2 Answers
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    2 Answers
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    active

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    up vote
    1
    down vote













    Note: I solve it base on that a,b are positive numbers(not integers)



    suppose $s=a+b$



    Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$



    Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$



    $Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$



    $0<s<=2794$



    for one solution, $0<a<s$



    (1) $0<3s^2+6s-sqrtDelta<2s(6+3s)$



    L:$3s^2+6s>sqrtDelta$



    $(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$



    $4(6+3s)(s^3-697s^2)>0$



    $0<s<697$



    R:$-3s^2-6s<sqrtDelta$ It is always true



    (2) $0<3s^2+6s+sqrtDelta<2s(6+3s)$



    L: always true



    R: $3s^2+6s>sqrtDelta$



    It is true for $0<s<697$



    Overall $a+b=s=1,2,...695$ or $696$



    $sum=242556$






    share|cite|improve this answer























    • @RossMillikan $a$ and $b$ can be noninteger
      – Mira from Earth
      2 days ago










    • I now see that.
      – Ross Millikan
      2 days ago














    up vote
    1
    down vote













    Note: I solve it base on that a,b are positive numbers(not integers)



    suppose $s=a+b$



    Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$



    Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$



    $Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$



    $0<s<=2794$



    for one solution, $0<a<s$



    (1) $0<3s^2+6s-sqrtDelta<2s(6+3s)$



    L:$3s^2+6s>sqrtDelta$



    $(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$



    $4(6+3s)(s^3-697s^2)>0$



    $0<s<697$



    R:$-3s^2-6s<sqrtDelta$ It is always true



    (2) $0<3s^2+6s+sqrtDelta<2s(6+3s)$



    L: always true



    R: $3s^2+6s>sqrtDelta$



    It is true for $0<s<697$



    Overall $a+b=s=1,2,...695$ or $696$



    $sum=242556$






    share|cite|improve this answer























    • @RossMillikan $a$ and $b$ can be noninteger
      – Mira from Earth
      2 days ago










    • I now see that.
      – Ross Millikan
      2 days ago












    up vote
    1
    down vote










    up vote
    1
    down vote









    Note: I solve it base on that a,b are positive numbers(not integers)



    suppose $s=a+b$



    Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$



    Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$



    $Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$



    $0<s<=2794$



    for one solution, $0<a<s$



    (1) $0<3s^2+6s-sqrtDelta<2s(6+3s)$



    L:$3s^2+6s>sqrtDelta$



    $(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$



    $4(6+3s)(s^3-697s^2)>0$



    $0<s<697$



    R:$-3s^2-6s<sqrtDelta$ It is always true



    (2) $0<3s^2+6s+sqrtDelta<2s(6+3s)$



    L: always true



    R: $3s^2+6s>sqrtDelta$



    It is true for $0<s<697$



    Overall $a+b=s=1,2,...695$ or $696$



    $sum=242556$






    share|cite|improve this answer















    Note: I solve it base on that a,b are positive numbers(not integers)



    suppose $s=a+b$



    Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$



    Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$



    $Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$



    $0<s<=2794$



    for one solution, $0<a<s$



    (1) $0<3s^2+6s-sqrtDelta<2s(6+3s)$



    L:$3s^2+6s>sqrtDelta$



    $(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$



    $4(6+3s)(s^3-697s^2)>0$



    $0<s<697$



    R:$-3s^2-6s<sqrtDelta$ It is always true



    (2) $0<3s^2+6s+sqrtDelta<2s(6+3s)$



    L: always true



    R: $3s^2+6s>sqrtDelta$



    It is true for $0<s<697$



    Overall $a+b=s=1,2,...695$ or $696$



    $sum=242556$







    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago


























    answered 2 days ago









    Mira from Earth

    1076




    1076











    • @RossMillikan $a$ and $b$ can be noninteger
      – Mira from Earth
      2 days ago










    • I now see that.
      – Ross Millikan
      2 days ago
















    • @RossMillikan $a$ and $b$ can be noninteger
      – Mira from Earth
      2 days ago










    • I now see that.
      – Ross Millikan
      2 days ago















    @RossMillikan $a$ and $b$ can be noninteger
    – Mira from Earth
    2 days ago




    @RossMillikan $a$ and $b$ can be noninteger
    – Mira from Earth
    2 days ago












    I now see that.
    – Ross Millikan
    2 days ago




    I now see that.
    – Ross Millikan
    2 days ago










    up vote
    0
    down vote













    Here is a start, not a full answer.



    Let $s=a+b$ and $p=ab$. Then $3 p (s + 2) = (s - 697) s^2$ as WA tells us.



    Then $s+2$ divides $(s - 697) s^2$ and so $s+2$ divides $2796 = 2^2 cdot 3 cdot 233$.






    share|cite|improve this answer



























      up vote
      0
      down vote













      Here is a start, not a full answer.



      Let $s=a+b$ and $p=ab$. Then $3 p (s + 2) = (s - 697) s^2$ as WA tells us.



      Then $s+2$ divides $(s - 697) s^2$ and so $s+2$ divides $2796 = 2^2 cdot 3 cdot 233$.






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Here is a start, not a full answer.



        Let $s=a+b$ and $p=ab$. Then $3 p (s + 2) = (s - 697) s^2$ as WA tells us.



        Then $s+2$ divides $(s - 697) s^2$ and so $s+2$ divides $2796 = 2^2 cdot 3 cdot 233$.






        share|cite|improve this answer















        Here is a start, not a full answer.



        Let $s=a+b$ and $p=ab$. Then $3 p (s + 2) = (s - 697) s^2$ as WA tells us.



        Then $s+2$ divides $(s - 697) s^2$ and so $s+2$ divides $2796 = 2^2 cdot 3 cdot 233$.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago


























        answered 2 days ago









        lhf

        155k9160364




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