Prove a square can't be written $5x+ 3$, for all integers $x$. Homework question, should I use induction?.. 
Help please
 A: Hint $\ $ Let $\,n\,$ by any integer. $\ {\rm mod}\ 5\!:\ n \equiv 0,\pm1,\pm2\,\Rightarrow\, n^2\equiv 0,1,4,\,$ so  $\ n^2\not\equiv 3\!\pmod{\!5}$
Remark $ $ It's easier if you know $\mu$Fermat: $\ 3 \equiv n^2\overset{\rm square}\Rightarrow\color{#0a0}{-1}\equiv \,n^4\ [\,\equiv \color{#c00}1$ by $\mu$Fermat]
This is a special case of Euler's Criterion: $\,{\rm mod}\ p=5\!:\,\ 3 \not\equiv n^2\, $ by $\ 3^{(p-1)/2}\equiv\color{#0a0}{-1}\not\equiv \color{#c00}1$
A: What are the only possibilities for the last digit of $5x + 3$ ?
Squares can only have as last digit: $0 , \ 1 , \ 4 , \ 5 , \ 6 , \ 9$.
A: Let $ A=\{y\in\mathbb{Z}: y=5x+3, x\in\mathbb{R} \}$
And, $y\in A \Rightarrow y$ are a multiple of 5 plus 3! The set of numbers that are multiples of 5 are $ \{...,-10,-5,0,5,10,...\} = \{n5: n \mathbb{Z} \}$
Think about it...
A: Consider the squares under $\mod{10}$ 
$$1^2\equiv{1}\pmod{10}$$
$$2^2\equiv{4}\pmod{10}$$
$$3^2\equiv{9}\pmod{10}$$
$$4^2\equiv{6}\pmod{10}$$
$$5^2\equiv{5}\pmod{10}$$
$$...$$
$$9^2\equiv{1}\pmod{10}$$
You will then see that numbers that end in 3 can not written as squares, since the only numbers that can end squares are $0,1,4,5,6,9$...
