Cubic Equation with Integer Roots!

Suppose that the roots of the polynomial are ${ r }_{ 1 }{ ,r }_{ 2 }{ ,r }_{ 3 }$ . We then have ${ x }^{ 3 }-13x+a=(x-{ r }_{ 1 })(x-{ r }_{ 2 })(x-{ r }_{ 3 })$ , and by Vieta's Formulas, ${ r }_{ 1 }+{ r }_{ 2 }+{ r }_{ 3 }=0$ , and ${ r }_{ 1 }{ r }_{ 2 }+{ r }_{ 1 }{ r }_{ 3 }+{ r }_{ 2 }{ r }_{ 3 }=-13$ .

Squaring the first formula yields ${ { r }_{ 1 } }^{ 2 }+{ { r }_{ 2 } }^{ 2 }+{ { r }_{ 3 } }^{ 2 }+{ 2r }_{ 1 }{ r }_{ 2 }+2{ r }_{ 1 }{ r }_{ 3 }+{ 2r }_{ 2 }{ r }_{ 3 }=0$ , which we notice to be the sum of the squares of the roots plus twice the left side of the second formula. Subtracting that from both sides leaves ${ { r }_{ 1 } }^{ 2 }+{ { r }_{ 2 } }^{ 2 }+{ { r }_{ 3 } }^{ 2 }=(-2)({ r }_{ 1 }{ r }_{ 2 }+{ r }_{ 1 }{ r }_{ 3 }+{ r }_{ 2 }{ r }_{ 3 })=26$ .

Thus, each root must lie between -5 and 5, inclusive, and with a little bit of guess-and-check, it becomes clear that the roots must be -4, 3, and 1 , or 4, -3, and -1 .

Since a is the product of the roots, a = 12 or -12 , and the sum of the squares of those two numbers is $\boxed{288}$

Clearly if we try and split 13 into two numbers such that one of them at least is a perfect square then we can ensure integer roots. Like if →x^3 - 13x +a=0 → x^3 - 9x - 4x +a=0 → x (x-3)(x+3) -4(x-a/4)=0→ Setting a=12 and a = -12 , We get 3 integer roots. On the other hand if we factorise as→ x^3 - 4x -9x + a=0 → x (x-2)(x+2) -9 (x-a/9)=0, We never get integer roots. Check for ur self!..! So sum of squares of integers' a' = (12)^2 + (-12)^2 = 144 + 144=288. Ans. Please note:- Im a beginner in maths, especially number theory, so if anyone finds some mistake plz suggest me. Thank you.

Let the roots are a1 ,a2, a3 then a1+a2+a3=0 and ,a1a2 +.....=-13 solve this equation and a1 ,a2 and a3 that is -4,3,1 and 4,-3,-2 so a=-12 or 12 Finally 288 is our answer...broooo

Used TI-83 Plus.

$-21 ~STO ~A\downarrow..........................\downarrow~~means~ENTER~KEY\\ A+1~STO~A\downarrow\\ Put~y~=x^3-13x+A=0~~in~graphing~mode.\\ In~window~~~x:--~10~to~10,~~~~~y:-~-~0.01~to~0.01.\\ Press ~~GRAPH\\ \text{Does the graph intersect x-axis at all three integer x?}\\ \text{Yes. then note down that A by pressing A and ENTER.}\\ \color{#D61F06}{\text{Press ENTER ENTER GRAPH.}} \\ \text{You get the next graph with A=A+1.}\\ \text{If there is no three integral cuts, go to next as under.}\\ \color{#3D99F6}{\text{Press ENTER GRAPH}} \\ \text{You will note that when |A|>18 the graph cuts the x-axis only in one}\\ \text{ point. This indicates that there are three real roots only for |A|<19.}\\ So~STOP~ at ~A=19.$