Equation in natural numbers

[MM]

Solve in natural numbers the equation
[tex]x^{3}+6x^{2}-109x-223=11^{y}[/tex]

[broniran]

Let [tex]P(x)=x^3+6x^2-109x-223[/tex]. [tex]P(x)[/tex] is equivalent to [tex](x+2)^3-111x-227[/tex] so we have to find all [tex]x[/tex] satisfying [tex](x+2)^3\equiv x+7(mod 11)[/tex] and we easily derive that [tex]x\equiv 5(mod 11)[/tex] or [tex]x\equiv 2 (mod11)[/tex].So it's enough to check if [tex]P(2)[/tex] or [tex]P(5)[/tex] is divisible by 11 since [tex]a\equiv b(mod c)\Rightarrow P(a)\equiv P(b)(mod c)[/tex] for all polynomials with integral coeficients. But [tex]P(2)=-399\equiv -3(mod 11)[/tex] and [tex]P(5)=-493\equiv 2(mod 11)[/tex].

[MM]

What about x=9, y=1.

[MM]

Since there have been no solution for a long time I will show the main moment.
[tex]x^{3}+6x^{2}-109x-223=11^{y}[/tex] [tex]\Leftrightarrow[/tex] [tex]\left(x-9\right)\left(x+2\right)\left(x+13\right)=11^{y}-11[/tex] and the rest is obvious.

[perfectmath]

Since there have been no solution for a long time I will show the main moment.
[tex]x^{3}+6x^{2}-109x-223=11^{y}[/tex] [tex]\Leftrightarrow[/tex] [tex]\left(x-9\right)\left(x+2\right)\left(x+13\right)=11^{y}-11[/tex] and the rest is obvious.

And how to solve next ??

[MM]

Sorry for answering after so much time. So we get that [tex]11[/tex] divides [tex](x-9)(x+2)(x+13)[/tex]. [tex]11[/tex] is prime so it divides one of the expressions in the brackets. But [tex]11=(x+2)-(x-9)=(x+13)-(x+2)[/tex]. So if [tex]11[/tex] divides one of the brackets then it divides the other two. So the expression on the left is divisible by [tex]11^3[/tex]. Hence [tex]11^2[/tex] divides [tex]11^y-11[/tex]. But if [tex]y\ge2[/tex] then [tex]11^2[/tex] divides [tex]11^y[/tex] and if it divides [tex]11^y-11[/tex] then it should divide [tex]11[/tex] which is impossible. So [tex]y=1[/tex]. Substituting we find it is really a solution.