My solution : We suppose that is true. Then by contradiction:
$a^2-4b-2=0$
$a^2=4b+2$
$a=2(b+1/2) ^{0.5}$
then $(b+1/2)$ is fraction and rooted by $0.5$ so the square root of any fraction $+$ any-Integer will give fraction so then $a$ must be fraction, not an integer. Contradiction.
Then $a^2 -4b \neq 2$?
Is that a good proof?
Suppose this is true then $a^2=2+4b$
$\implies 2|a^2$
$\implies 2|a$
$\implies $ 4 is a multiple of L.H.S. but it is not a multiple of R.H.S.
which leads to a contradiction
$\implies 4((\frac{a}{2})^2-b)=2$
$\implies (\frac{a}{2})^2-b=\frac{1}{2}$
Now you can easily see this is a contradiction
– absolute friend Jun 17 '15 at 02:25Your proof is incorrect because you make the assertion that "$\sqrt{b+\frac{1}{2}}$ is a fraction so $2 \sqrt{b+\frac{1}{2}}$ is also a fraction." It may very well turn out to be a whole number! You will need some other kind of reasoning to get a contradiction.
no, I think your argument about the square root is not clear or proven. Also
How about
a^2=4b+2 Suppose a and b are integers, then 4b+2 is even, so a^2 is also an integer and a must be even. Let 2c=a, then we get 4c^2 = 4b + 2, c^2 - b = 1/2, we cannot have the difference of two integers being fractional, so we have a contradiction, hence a and b both cannot be integers. (sorry my earlier proof mistakenly tried to show either a or b were integers).
Suppose $a^2=4b+2 $. $a$ must be of the form $2c$ or $2c+1$ (i.e., even or odd).
If $a=2c$, then $2 = (2c)^2-4b = 4c^2-4b = 4(c^2-b) $. But 4 divides the right side but not the left, so this is impossible.
If $a=2c+1$, then $2 = (2c+1)^2-4b = 4c^2+4c+1-4b = 4(c^2+c-b)+1 $. But the left side is even and the right side is odd, so this is also impossible.
Note that this works, with minor changes, for $a^2 = 4b+3 $.
Suppose $a^2-4b=2$. Reduce both sides modulo 4 to obtain $a^2=2\mod{4}$. But $a\equiv \{-1,0,1,2\}$, so $a^2\equiv \{0,1\}$. Hence there is no solution.
