So my cousin is in the math team (7th grade) and he was asking me for help on one of his problems but I don't know how to solve

For what positive integer n does $3n^3 + 3n^2 + 4n = n^n$

anyone know how to do this?

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    $\begingroup$ $n=4........................$ $\endgroup$ – Will Jagy Apr 26 '14 at 5:59
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    $\begingroup$ Divide through by $n$, then show $n$ goes into 4 without remainder. $\endgroup$ – Gerry Myerson Apr 26 '14 at 6:02
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    $\begingroup$ $n^n$ grows very rapidly, and the left-hand side is even, so the work is quick. $\endgroup$ – André Nicolas Apr 26 '14 at 6:02
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    $\begingroup$ $4 = n(n^{n-2} - 3n -3 )$, and each side is an integer. $4$ only has $1,2,4$ as factors; $n=1$ and $n=2$ don't work. $\endgroup$ – user139388 Apr 26 '14 at 6:06
  • $\begingroup$ @user139388 You should write that up as an answer; it's an excellent one. $\endgroup$ – Steven Stadnicki Apr 26 '14 at 6:47

This is the solution I mentioned in the comments and was encouraged to put as a full post:

Factor $n$ out of the equation and rearrange to get $$ 4 = n(n^{n-2} - 3n - 3). $$ This is a factorization of $4$ into positive integers. The only positive integer factors of $4$ are $1,2$ and $4$, but $n=1$ and $n=2$ don't work, so $n=4$.


From an algebraic pont of view, you could consider the function: $$f(x)=3x^3 + 3x^2 + 4x- x^x$$ and notice that $f(0)=-1$, $f(1)=9$, $f(2)=40$, $f(3)=93$, $f(4)=0$. For $x \gt 4$, the term $x^x$ is very predominant and $f(x)$ decreases from $0$ to $-\infty$.

So, in the integer domain, there is only one solution which is $n=4$. For real solutions, there is another root between $0$ and $1$ (in fact $x=0.162884$).

  • $\begingroup$ Just be precise with the justification, grows very rapidly is not what is expected on a math contest(but does the work as an intuition and hint, like here) $\endgroup$ – chubakueno Apr 26 '14 at 6:17
  • $\begingroup$ @chubakueno. Could you rephrase for me, please ? I would really appreciate. Thanks and cheers. $\endgroup$ – Claude Leibovici Apr 26 '14 at 6:21
  • $\begingroup$ I would suggest the OP proving it with induction. If this was still not covered, divisibility arguments like the one of @user139388 work. Cheers too! $\endgroup$ – chubakueno Apr 26 '14 at 6:26

I would also try the general approach, draw both functions f(x) and g(x) with some graph software (like graph :-) ) and then check the solutions by inserting them into the equations. If the equation is rewritten like the user @user139388 did,then you can see the only positive integer solution is 4.

(I know that the general approach is to advanced for 7th grade,but it tells you what other possible solutions are and nowadays kids have no problems with the use of computers and simple computer programs like graph etc :-)) (below the graph there is a re-arranged equation and one can se that the only positive INTEGER solution is n=4)

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