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)
Hello,
First, put the expression in a more familiar form : why can't $\displaystyle 3n + 5$ ever be a perfect square ? For $\displaystyle 3n + 5$ to be a perfect square, we need to have : $\displaystyle 3n + 5 = a^2$ for some integer $\displaystyle a$, that is, $\displaystyle 3n = a^2 - 5$. Thus, saying that no integer a satisfies this condition is equivalent to showing that $\displaystyle a^2 - 5$ is never divisible by three. The "alternative" question is in fact necessary to solve the first question, so see the following :why can't 5+3n ever be a perfect square? (n is an integer)
Suppose $\displaystyle x$ is divisible by three, so $\displaystyle x^2$ is divisible by three, so obviously $\displaystyle x^2 - 5$ is not divisible by three ($\displaystyle x^2 - 5 \equiv 1 \pmod{3}$).why isn't x^2-5 divisible by three? (x is an integer)
Suppose x is not divisible by three, so $\displaystyle x^2 \neq 0 \pmod{3}$. Therefore, $\displaystyle x^2 - 5 \neq -5 \pmod{3}$, that is, $\displaystyle x^2 - 5 \neq 1 \pmod{3}$, or, pushing even further, $\displaystyle (x^2 \ mod 3) - 1 \neq 1 \pmod{3}$. Note that squares can only be equal to $\displaystyle 0$ or $\displaystyle 1$ modulo 3, but we already considered the case when it is equal to 0 modulo 3 (it is divisible by three). So assume $\displaystyle x^2 \equiv 1 \pmod{3}$, so $\displaystyle x^2 - 5 \equiv -4 \equiv 2 \pmod{3}$.
Conclusion : $\displaystyle x^2 - 5$ is either equal to 1 or 2 modulo 3, and thus cannot be divisible by three.
Final conclusion : $\displaystyle 3n + 5$ cannot be a perfect square.
Does that make sense ?
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