Problem of the Week # 275 - Aug 07, 2017

[Ackbach]

Here is this week's POTW:

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Prove that there are unique positive integers $a$, $n$ such that $a^{n+1}-(a+1)^n=2001$.

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[Ackbach]

Re: Problem Of The Week # 275 - Aug 07, 2017

This was Problem A-5 in the 2001 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution to this week's POTW, which follows:

[sp]Expand $(a+1)^n$ binomially to get $$a^{n+1} - (a+1)^n = a^{n+1} - a^n - {\textstyle {n\choose1}}a^{n-1} - \ldots - {\textstyle {n\choose n-1}}a - 1,$$ from which $$a(a^n - a^{n-1} - na^{n-2} - \ldots - n) = 2001+1 = 2002.$$ Therefore $a$ is a factor of $2002 = 2*7*11*13.$ On the other hand, a similar argument shows that if $b=a+1$ then $$a^{n+1} - (a+1)^n = (b-1)^{n+1} - b^n = b\bigl(b^n - {\textstyle {n+1\choose1}}b^{n-1} + \ldots + (-1)^n - b^{n-1}\bigr) + (-1)^{n+1} = 2001,$$ so that $b$ is a factor of 2002 (if $n$ is even) or 2000 (if $n$ is odd).

So there are two cases to consider: either (1) $a$ and $a+1$ are both factors of $2002$, or (2) $n$ is odd, $a$ is a factor of $2002$ and $a+1$ is a factor of $2000$.

For case (1), the only consecutive factors of $2002$ are $13$ and $14$. With $n=2$, that leads to the solution $13^3 - 14^2 = 2197 - 196 = 2001.$

As $x$ increases from $0$, the function $f_{13}(x) = 13^{x+1} - 14^x$ increases to a maximum and then decreases (very rapidly). It takes the value $2001$ when $x=2$, at which point it is still increasing. It takes the value $2001$ again when it is decreasing, and conceivably this might happen at an integer value of $x$. However, you can check that $f_{13}(x)$ reaches its maximum when $x$ is approximately $34.22$, and it has already decreased to $0$ when $x\approx 34.61$. So it cannot be equal to $2001$ at any integer value apart from $2$.

For case (2) there is again just one candidate, namely $a=7$ and $a+1 = 8$. But the function $f_7(x) = 7^{x+1} - 8^x$ satisfies $f_7(3) = 1889$, $f_7(4) = 12711$. So it does not take the value $2001$ at any integer point of its increasing stage. Its maximum occurs at approx. $14.07$ and it is zero at approx. $14.57$. So it cannot take the value $2001$ at an integer point during its decreasing stage either.

Thus the only solution of $a^{n+1} - (a+1)^n = 2001$ for positive integers is $a = 13$, $n=2$.[/sp]