Polynomial problem with unknown coefficients $a, b, c$

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$p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.



Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-fraca^2a+1$ and then getting $c=-fraca^4a+1$ does not help much I guess. Will someone tell me how I should be thinking about this question? Thanks.




algebra-precalculus elementary-number-theory polynomials




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  • Don't forget that $a,b,c$ are integers. I suppose that this implies conditions.
    – Claude Leibovici
    Jul 28 at 7:26














up vote
3
down vote

favorite












$p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.



Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-fraca^2a+1$ and then getting $c=-fraca^4a+1$ does not help much I guess. Will someone tell me how I should be thinking about this question? Thanks.




algebra-precalculus elementary-number-theory polynomials




share|cite|improve this question





















  • Don't forget that $a,b,c$ are integers. I suppose that this implies conditions.
    – Claude Leibovici
    Jul 28 at 7:26












up vote
3
down vote

favorite









up vote
3
down vote

favorite











$p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.



Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-fraca^2a+1$ and then getting $c=-fraca^4a+1$ does not help much I guess. Will someone tell me how I should be thinking about this question? Thanks.




algebra-precalculus elementary-number-theory polynomials




share|cite|improve this question













$p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.



Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$,
So $(a-b)[a(a+b)+b]=0$, so $b=-fraca^2a+1$ and then getting $c=-fraca^4a+1$ does not help much I guess. Will someone tell me how I should be thinking about this question? Thanks.





algebra-precalculus elementary-number-theory polynomials





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edited Jul 28 at 8:53









greedoid

26.1k93473




26.1k93473









asked Jul 28 at 7:14









shgdh fgxbcv

1407




1407











  • Don't forget that $a,b,c$ are integers. I suppose that this implies conditions.
    – Claude Leibovici
    Jul 28 at 7:26
















  • Don't forget that $a,b,c$ are integers. I suppose that this implies conditions.
    – Claude Leibovici
    Jul 28 at 7:26















Don't forget that $a,b,c$ are integers. I suppose that this implies conditions.
– Claude Leibovici
Jul 28 at 7:26




Don't forget that $a,b,c$ are integers. I suppose that this implies conditions.
– Claude Leibovici
Jul 28 at 7:26










2 Answers
2






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2
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accepted










Put $n=a+1$, then: $$b=-a^2over a+1 = -(n-1)^2over n = -n^2-2n+1over n = -n+2-1over n inmathbbZ$$



So $$nmid 1implies n =pm 1 implies a=0 vee a= -2 $$
Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.






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  • Ah i see, thanks!
    – shgdh fgxbcv
    Jul 28 at 11:37

















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1
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Since $b=-fraca^2a+1$, we see that
$$c=-a^3-ab=-a^3+fraca^3a+1=-fraca^4a+1$$ and since $gcd(a^4,a+1)=1,$ we obtain:



$a+1in-1,1$ and since $aneq0$, we obtain $a=-2$, $c=16$, $b=4$,
$$f(x)=x^3-2x^2+4x+16$$ and
$$f(13)=13^3-2cdot13^2+4cdot13+16=1927.$$






share|cite|improve this answer





















  • ok it's slightly similar to the solution above so thanks!
    – shgdh fgxbcv
    Jul 28 at 11:38










  • @shgdh fgxbcv I think my step with $gcd$ is better.
    – Michael Rozenberg
    Jul 28 at 12:27










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Put $n=a+1$, then: $$b=-a^2over a+1 = -(n-1)^2over n = -n^2-2n+1over n = -n+2-1over n inmathbbZ$$



So $$nmid 1implies n =pm 1 implies a=0 vee a= -2 $$
Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.






share|cite|improve this answer























  • Ah i see, thanks!
    – shgdh fgxbcv
    Jul 28 at 11:37














up vote
2
down vote



accepted










Put $n=a+1$, then: $$b=-a^2over a+1 = -(n-1)^2over n = -n^2-2n+1over n = -n+2-1over n inmathbbZ$$



So $$nmid 1implies n =pm 1 implies a=0 vee a= -2 $$
Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.






share|cite|improve this answer























  • Ah i see, thanks!
    – shgdh fgxbcv
    Jul 28 at 11:37












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Put $n=a+1$, then: $$b=-a^2over a+1 = -(n-1)^2over n = -n^2-2n+1over n = -n+2-1over n inmathbbZ$$



So $$nmid 1implies n =pm 1 implies a=0 vee a= -2 $$
Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.






share|cite|improve this answer















Put $n=a+1$, then: $$b=-a^2over a+1 = -(n-1)^2over n = -n^2-2n+1over n = -n+2-1over n inmathbbZ$$



So $$nmid 1implies n =pm 1 implies a=0 vee a= -2 $$
Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 28 at 7:41


























answered Jul 28 at 7:33









greedoid

26.1k93473




26.1k93473











  • Ah i see, thanks!
    – shgdh fgxbcv
    Jul 28 at 11:37
















  • Ah i see, thanks!
    – shgdh fgxbcv
    Jul 28 at 11:37















Ah i see, thanks!
– shgdh fgxbcv
Jul 28 at 11:37




Ah i see, thanks!
– shgdh fgxbcv
Jul 28 at 11:37










up vote
1
down vote













Since $b=-fraca^2a+1$, we see that
$$c=-a^3-ab=-a^3+fraca^3a+1=-fraca^4a+1$$ and since $gcd(a^4,a+1)=1,$ we obtain:



$a+1in-1,1$ and since $aneq0$, we obtain $a=-2$, $c=16$, $b=4$,
$$f(x)=x^3-2x^2+4x+16$$ and
$$f(13)=13^3-2cdot13^2+4cdot13+16=1927.$$






share|cite|improve this answer





















  • ok it's slightly similar to the solution above so thanks!
    – shgdh fgxbcv
    Jul 28 at 11:38










  • @shgdh fgxbcv I think my step with $gcd$ is better.
    – Michael Rozenberg
    Jul 28 at 12:27














up vote
1
down vote













Since $b=-fraca^2a+1$, we see that
$$c=-a^3-ab=-a^3+fraca^3a+1=-fraca^4a+1$$ and since $gcd(a^4,a+1)=1,$ we obtain:



$a+1in-1,1$ and since $aneq0$, we obtain $a=-2$, $c=16$, $b=4$,
$$f(x)=x^3-2x^2+4x+16$$ and
$$f(13)=13^3-2cdot13^2+4cdot13+16=1927.$$






share|cite|improve this answer





















  • ok it's slightly similar to the solution above so thanks!
    – shgdh fgxbcv
    Jul 28 at 11:38










  • @shgdh fgxbcv I think my step with $gcd$ is better.
    – Michael Rozenberg
    Jul 28 at 12:27












up vote
1
down vote










up vote
1
down vote









Since $b=-fraca^2a+1$, we see that
$$c=-a^3-ab=-a^3+fraca^3a+1=-fraca^4a+1$$ and since $gcd(a^4,a+1)=1,$ we obtain:



$a+1in-1,1$ and since $aneq0$, we obtain $a=-2$, $c=16$, $b=4$,
$$f(x)=x^3-2x^2+4x+16$$ and
$$f(13)=13^3-2cdot13^2+4cdot13+16=1927.$$






share|cite|improve this answer













Since $b=-fraca^2a+1$, we see that
$$c=-a^3-ab=-a^3+fraca^3a+1=-fraca^4a+1$$ and since $gcd(a^4,a+1)=1,$ we obtain:



$a+1in-1,1$ and since $aneq0$, we obtain $a=-2$, $c=16$, $b=4$,
$$f(x)=x^3-2x^2+4x+16$$ and
$$f(13)=13^3-2cdot13^2+4cdot13+16=1927.$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 28 at 8:34









Michael Rozenberg

87.7k1578180




87.7k1578180











  • ok it's slightly similar to the solution above so thanks!
    – shgdh fgxbcv
    Jul 28 at 11:38










  • @shgdh fgxbcv I think my step with $gcd$ is better.
    – Michael Rozenberg
    Jul 28 at 12:27
















  • ok it's slightly similar to the solution above so thanks!
    – shgdh fgxbcv
    Jul 28 at 11:38










  • @shgdh fgxbcv I think my step with $gcd$ is better.
    – Michael Rozenberg
    Jul 28 at 12:27















ok it's slightly similar to the solution above so thanks!
– shgdh fgxbcv
Jul 28 at 11:38




ok it's slightly similar to the solution above so thanks!
– shgdh fgxbcv
Jul 28 at 11:38












@shgdh fgxbcv I think my step with $gcd$ is better.
– Michael Rozenberg
Jul 28 at 12:27




@shgdh fgxbcv I think my step with $gcd$ is better.
– Michael Rozenberg
Jul 28 at 12:27












 

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