Suppose that $a,p$ are nonnegative integers such that $p$ is prime and $p\nmid (a-1)$. If $(a-1)^{p+1}+a^p=(a+1)^{p-1}$, find the sum of all possible values of $a$.

We can't have $p = 2$ since the equation $$(a-1)^3+a^2 = (a+1)$$ has no integer solutions. We now take two cases:

Case 1: $p \mid (a+1)$

In this case, $a \equiv -1 \pmod{p}$. Then $(a-1)^{p+1} \equiv (-2)^{p+1} \equiv 4 \equiv -a^p \equiv -a \pmod{p}$. Thus $a \equiv 4 \pmod{p}$ and therefore $p \mid 3$, so that $p = 3$. Thus $a = 2$ in this case.

Case 2: $p \nmid (a+1)$

We have $$(a-1)^{p+1}+a^p \equiv (a-1)^2+a^p \equiv 1 \pmod{p}.$$ If $a \not \equiv 0 \pmod{p}$, then we have $(a-1)^2+a \equiv 1 \pmod{p}$, which gives $a(a-1) \equiv 0 \pmod{p}$. Thus $p \mid (a-1)$, a contradiction. Thus $a \equiv 0 \pmod{p}$.

How do I continue from here?

  • 3
    $\begingroup$ I would think that the LHS would be greater than the RHS for almost any value of a and p .. certainly for a>2 and p>2 ... so couldn't you narrow it down fairly quickly that way, i.e. prove that for most values LHS > RHS, and see what's left? $\endgroup$
    – Bram28
    Jun 2, 2017 at 17:48

2 Answers 2


Hint $$1 \geq \frac{(a-1)^{p+1}}{(a+1)^{p-1}}=(a-1)^2 (1-\frac{2}{a+1})^{p-1}\geq (a-1)^2 (1-\frac{2(p-1)}{a+1})$$ with the last inequality following by Bernoulli.

Thus $$a+1 \geq (a-1)^2(a+3-2p) \\ 2 =a+1-(a-1)\geq a+1-(a-1)^2\geq (a-1)^2(a+2-2p)$$

This implies that either $(a+2-2p) \leq 0$ or $a-1=1$ and $(a+2-2p) \in \{ 1,2 \}$.

Combine this with $p|a$ to complete the proof.

Added If $a=p \geq 3$ then $$(p-1)^{p+1}+p^p=(p+1)^{p-1}$$

and hence $$(p-1)^{p+1}+p^p=(p+1)^{p-1} \pmod{p^3} \\ \binom{p+1}{2}p^2(-1)^{p-1}+\binom{p+1}{1}p(-1)^p+(-1)^{p+1}=\binom{p-1}{2}p^2+\binom{p-1}{1}p+1\pmod{p^3} \\ \frac{(p+1)p}{2}p^2-(p+1)p+1= \frac{(p-1)(p-2)}{2}p^2+(p-1)p+1 \pmod{p^3} \\ -p^2-p+1= p^2+p^2-p+1 \pmod{p^3} \\ 3p^2=0 \pmod{p^3} \\ $$

This shows $p=3$. The case $p=2$ is easy to study separately.

  • $\begingroup$ How do we deal with the case that $a = p$? $\endgroup$ Jun 3, 2017 at 0:50
  • $\begingroup$ @user19405892 Check the added part. $\mod{p^2}$ was not enough that's why I went $\pmod{p^3}$. $\endgroup$
    – N. S.
    Jun 3, 2017 at 1:09

I will assume $p$ is odd.

Consider the monic polynomial $f(X) = (X-1)^{p+1} + X^p - (X+1)^{p-1}$.

It has $0$ as a root, so dividing by $X$ we find $g(X) = f(X)/X$ is another monic polynomial.

The constant term of $g$ is ${p+1 \choose p}(-1)^p-{p-1 \choose p-2} = (p+1)(-1)^p-(p-1) = -2p$.

The rational root theorem gives us the possible integer roots of $f$ as $0, \pm 1, \pm 2, \pm p, \pm 2p.$

We should be able to test each of these candidates and thus find all integer roots, and their sum.

Since we are interested only in positive integers (and $0$ would not affect the sum) we only need to check $1, 2, p, 2p$.

$a = 1$ is impossible because $f(1) = 1 - 2^{p-1} = 0$ requires $p = 1$ which is not prime.

$a = 2$ gives the equation $2^p + 1 = 3^{p-1}$ which is true only when $p = 3$.

$a = p$ leads to $f(p) = (p-1)^{p+1} + p^p - (p+1)^{p-1}=0$.

But $f(p) \geq (p-1)^{p-1}\left((p-1)^2+(p-1)-\left(1+\frac{2}{p-1}\right)^{p-1}\right)\geq 4^4(4^2+4-e^2) \gt 0$ for $p \geq 5.$ We can check $p = 3$ individually to find $f(3) = 27 \gt 0$ so $p$ is not a root.

[A much simpler way occurred to me, since $p$ is odd this is impossible because $(p-1)^{p+1} + p^p - (p+1)^{p-1}$ is odd too].

$a = 2p$ should work similarly as well.


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