1

So I am solving a number theory problem and as the last step I have to figure out all perfect squares in the form of $103k-104$ where $k$ is an integer.

I am stuck, is there any hint?

Mr.Infinity
  • 69

2 Answers2

2

If there is a perfect square of the form $103k-104$, then there is a square $q^2$ satisfying $$q^2 \equiv_{103} -1.$$ So $-1$ would have to be a square in $\mathbb{Z}/103\mathbb{Z}$. But as $103$ is prime, this would imply $103$ must be $1 \pmod 4$, which is not true. [Do you see why this is.]

So there are no such squares.

Mike
  • 23,484
  • Because the order of q mod p would be 4 and therefore 4 would divide the order of that group, which is 103 ? – Thomas Jun 06 '21 at 20:40
  • The order of that group is $102$. Since $4$ does not divide $102$, we get a contradiction. – Angelo Jun 06 '21 at 20:43
  • @Angelo Ah yes it is the multiplication group without the 0 class not the additive one, so 102 elements... thanks – Thomas Jun 06 '21 at 20:47
  • 1
    Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 06 '21 at 20:53
1

Infinity, I thought about your problem and this was what I got, I hope it's useful!

if $ k\in\mathbb{Z^+} $ in order to $ 103k-104$ be a perfect square $q$ , at first it needs to positive. So I defined $\,m = k + 2,\;$where $\;m\in \mathbb{Z^+}$, therefore, it's possible to rewrite your equation as it follows:

$$103(m+2)-104 =q^2 \Leftrightarrow 102+103m=q^2 $$ by doing so, I tried to complete the square

knowing that:$$(a+b)^2=a^2+b^2+2ab$$

if there is a m such that: $(b+m)^2=q^2$

then we would have: $$102+103m = (b+m)^2 \implies \\b^2 = 102,\\ m(2b+m)= 103m$$

and solving this equation, I got: $m=103-2\sqrt{102}$ which $\in \mathbb{R}-\mathbb{Q}$

But we said earlier that, $m \in \mathbb{Z}$ , so we are in a absurd!

Thank you for the question, I hope I've helped somehow

Angelo
  • 13,952
Ramon Coche
  • 36
  • Actually

    $102+103m=(b+m)^2\implies\begin{cases}b^2=102+\lambda m\m(2b+m)=103m-\lambda m\end{cases}$

    where $;\lambda\in\mathbb{Z};.$

     – Angelo Jun 06 '21 at 20:35
  • You need to get an absurd even though $;\lambda\ne0;.$ – Angelo Jun 06 '21 at 20:39