0
$\begingroup$

Find all primes $p$ for which $2017^{p-1}+p^3$ is a perfect square.

Note that $p \neq 2017$ and let $2017^{p-1}+p^3 = k^2$ for some integer $k$. By Fermat's Little Theorem, $2017^{p-1} \equiv 1 \pmod{p}$ and so $$2017^{p-1}+p^3 \equiv 1 \pmod{p}.$$ Thus $k = \pm1+pm$ for some integer $m$. If $k = 1+pm$, then $$2017^{p-1} = 1+2pm+p^2m^2-p^3 = 1+p(2m+pm^2-p^2)$$ and therefore $$2017^{p-1}-1 = p(2m+pm^2-p^2).$$

I didn't see how to continue from here.

  • 0
    Where is this problem from? Tasks that involve the current year are often from contests or exams, and we're wary of providing answers to them while they may still be ongoing. – 2017-01-22

1 Answers 1

2

Note $p=2$ is a solution.

If $p$ is odd,

$$2017^{p-1}+p^3 = k^2 \implies p^3=(k-2017^{\frac{p-1}{2}})(k+2017^{\frac{p-1}{2}})$$ Since $k-2017^{\frac{p-1}{2}}

However, note that the difference between $(k-2017^{\frac{p-1}{2}}$ and $k+2017^{\frac{p-1}{2}}$ is larger than $p^3-1$ $p^2-p$ for all $p$.

So we have that it is impossible when $p \equiv 1 \pmod{2}$.

So the answer is $p=2$.

  • 1
    Can you show how to prove that $2 \cdot 2017^{\frac{p-1}{2}} > p^3-1$? – 2017-01-22
  • 0
    @user19405892 Well, you first put $p=2k+1$. So we have to prove that $$2 \times 2017^{k} > 8k^3+12k^2+6k \implies 2017^{k}>4k^3+6k^2+3k$$ – 2017-01-22
  • 0
    Do we then prove it by induction? – 2017-01-22
  • 0
    @user19405892 Then you can prove (using calculus) that for all real $x>0$ $$2017^x >4x^3+6x^2+3x$$ – 2017-01-22
  • 0
    @user19405892 Well, you can also prove it using induction. – 2017-01-22
  • 0
    @user19405892 $2017>e^7$, $e^x >x+1$ $\Rightarrow$ $$2017^x \ge e^{7x} = (e^x)^7 \ge (x+1)^7 >4x^3+6x^2+3x$$ – 2017-01-22
  • 0
    @user19405892 Do you understand? – 2017-01-22
  • 0
    No, can you show using the induction? – 2017-01-22
  • 0
    @user19405892 Yes, you can.It is true for $x=0$. Assume it is true for $n$. Then for $n+1$, $$2017^{x+1} = 2017 \times 2017^x \ge 2017 \times (4x^3+6x^2+2x) >4(x+1)^3+6(x+1)^2+3(x+1)$$ The last part follows from direct calculation. – 2017-01-22
  • 0
    @user19405892 Do you understand? – 2017-01-22