3
$\begingroup$

Does anyone know how to solve Diophantine equation: $${\sqrt{n}}^\sqrt{n}-11 =m!^2.$$

I tried to substitute $\sqrt{n}=k$ then equation becomes $$k^k-11=m!^2\\\implies k^k=m!^2+11=(m!-1)(m!+1)+12$$ which means suppose $m\geq2$ then $\gcd{(m!-1,m!+1)}=1$. Does this give any hint? I could think upto here only. Please help.

  • 1
    It should help that after $m=9$, the number $m!^2 + 11$ will end in "11". – 2017-01-24
  • 2
    $n$ must be a perfect square. Otherwise $\sqrt n^{\sqrt n}$ would be trascendental. Together with Aston's comment and the fact that no square can end with $11$, this implies that $n$ must be an odd perfect square. – 2017-01-24
  • 0
    $k^k\equiv (m!+1)(m!-1)\pmod 4$. Then $k^k\equiv-1\pmod 4$. So, $k\equiv -1\pmod 4$. – 2017-01-25
  • 0
    What is $n$ here? – 2017-01-25
  • 0
    @ajotatxe: We want $k^k$ to end with $11$, not the square. – 2017-01-25
  • 0
    $k$ must be odd because all the factorials aside from $1$ are even. I would try $k=1,3,5,7$ and report failure. $3^3-11$ gives a square, but $4$ is not a factorial. Then squares are so rare it won't happen. – 2017-01-25
  • 1
    If $m \geq 11$, then $11 \mid m!$ and so $11 \| m!^2+11$. It follows that $11 \| k^k$, a contradiction. – 2017-01-25
  • 0
    what does || symbol mean? – 2017-01-25
  • 0
    Exactly divides. – 2017-01-25

1 Answers 1

1

Let $$k^k-11 = (m!)^2.$$ If $k>=11$ then $$121\not|\, LHS,\quad 121\,|\, RHS,\quad LHS\not= RHS.$$ In the other hand, $k$ is odd and $k^k > 11$, so $$k\in\{3,5,7,9\}.$$ Note than: $$\sqrt{3^3-11} = 4 \not= m!,$$ $$\sqrt{5^5-11} = \sqrt{3114}\in(55,56),\quad \sqrt{5^5-11}\not\in\mathbb N,$$ $$\sqrt{7^7-11} = \sqrt{823532}\in(907,908),\quad \sqrt{7^7-11}\not\in\mathbb N,$$ $$\sqrt{9^9-11} = \sqrt{387420478}= \sqrt{11683^2-1}\in(11682,11683),\quad \sqrt{9^9-11}\not\in\mathbb N.$$

So the issue diophantine equation has not solutions.