0
$\begingroup$

If $a, b$ are non-square whole numbers, and $c$ is an positive whole number, prove there exists no solution to the following equation:

$$\sqrt{a}+\sqrt{b}=c$$

I have absolutely no idea where to start with this question as the possibilities of $a,b,c$ is too broad for what I am used to.

$\endgroup$
4
  • 4
    $\begingroup$ Over a thousand rep points and you don't even abide by the guidelines for asking a question?? What have you tried? $\endgroup$
    – jameselmore
    Commented Nov 30, 2015 at 4:45
  • $\begingroup$ @jameselmore Sorry, bad habit of mine $\endgroup$
    – user285523
    Commented Nov 30, 2015 at 4:47
  • $\begingroup$ Hint: could you answer this question if you knew that $\sqrt{a}-\sqrt{b}$ was rational? If so, how could you show this? $\endgroup$
    – Alex Wertheim
    Commented Nov 30, 2015 at 4:48
  • 1
    $\begingroup$ $(\sqrt a)^2 = (c-\sqrt b)^2$. Now equate the coefficients of $\sqrt b$. $\endgroup$
    – Christopher Carl Heckman
    Commented Nov 30, 2015 at 4:54

1 Answer 1

Reset to default
2
$\begingroup$

Suppose a solution exists. Then

$$\sqrt a = c-\sqrt b$$ $$a = c^2 + b -2\sqrt{b}$$ $$\sqrt b = \frac{c^2+b-a}2$$ $$b=\left(\frac{c^2+b-a}2\right)^2$$

Which shows that either b is the square of some non-integer (if $c^2+b-a$ is odd), i.e. b is not an integer, or that b is the square of an integer (if $c^2+b-a$ is even). Both cases lead to contradictions. Hence, there is no solution.

$\endgroup$

You must log in to answer this question.