Thread: Why can't I be a perfect square?

1. Why can't I be a perfect square?

why can't 5+3n ever be a perfect square? (n is an integer)

alternatively.

why isn't x^2-5 divisible by three? (x is an integer)

2. Hello,
why can't 5+3n ever be a perfect square? (n is an integer)
First, put the expression in a more familiar form : why can't $3n + 5$ ever be a perfect square ? For $3n + 5$ to be a perfect square, we need to have : $3n + 5 = a^2$ for some integer $a$, that is, $3n = a^2 - 5$. Thus, saying that no integer a satisfies this condition is equivalent to showing that $a^2 - 5$ is never divisible by three. The "alternative" question is in fact necessary to solve the first question, so see the following :

why isn't x^2-5 divisible by three? (x is an integer)
Suppose $x$ is divisible by three, so $x^2$ is divisible by three, so obviously $x^2 - 5$ is not divisible by three ( $x^2 - 5 \equiv 1 \pmod{3}$).
Suppose x is not divisible by three, so $x^2 \neq 0 \pmod{3}$. Therefore, $x^2 - 5 \neq -5 \pmod{3}$, that is, $x^2 - 5 \neq 1 \pmod{3}$, or, pushing even further, $(x^2 \ mod 3) - 1 \neq 1 \pmod{3}$. Note that squares can only be equal to $0$ or $1$ modulo 3, but we already considered the case when it is equal to 0 modulo 3 (it is divisible by three). So assume $x^2 \equiv 1 \pmod{3}$, so $x^2 - 5 \equiv -4 \equiv 2 \pmod{3}$.

Conclusion : $x^2 - 5$ is either equal to 1 or 2 modulo 3, and thus cannot be divisible by three.

Final conclusion : $3n + 5$ cannot be a perfect square.

Does that make sense ?

3. thanks! I need to find out what mod / modulo are now

4. Originally Posted by birdmw
thanks! I need to find out what mod / modulo are now
It's an awesome tool to study divisibility and modular arithmetic in general. Some links :
Modular arithmetic - Wikipedia, the free encyclopedia
Math Forum - Ask Dr. Math

They really make problem solving quicker and easier, and give a steady working ground in number theory word problems