What is the condition for $ ( \sqrt{a} - b )( \sqrt{a} + b ) $ to be 1?

Question

I think its not a good level question, but I still have problem that what are the criteria for any two positive reals $ a , b $ so that, $$ ( \sqrt{a} - b )( \sqrt{a} + b ) = 1 $$

I need to know what type of numbers $ a $ and $ b $ need to be in order to satisfy the above equation.

(For instance: $ a $ should be prime, etc.)

I found:

• When any of $ a $ or $ b $ is equal to 1, this doesn't holds.

• When $ a $ is composite, then also ths doesn't holds.

Answer 1

Note that $$(\sqrt{a}-b)(\sqrt{a}+b)=a+\sqrt{a}b-b\sqrt{a}-b^2=a-b^2.$$ So this will equal $1$ iff $a=b^2+1$. In particular, for each $b\in\mathbb{R}$, there is a unique $a\in\mathbb{R}$ which makes it true. A number $a$ has the form $b^2+1$ for some $b\in \mathbb{R}$ iff $a\geq 1$ (in which case $b=\sqrt{a-1}$).

If you want to require $a$ and $b$ to be integers, the same condition holds: you need $a=b^2+1$. There is again a unique choice of $a$ for each choice of $b$. The possible values of $a$ are just all integers which are one more than a perfect square. Note that such an $a$ certainly doesn't have to be prime: indeed, if $b>1$ is odd, then $b^2+1$ is even and greater than $2$ and therefore is never prime.

Answer 2

Hint: $( \sqrt{a} - b )( \sqrt{a} + b ) =a-b^2$.

Answer 3

$$(\sqrt a - b)(\sqrt a + b) = a -b^2=1 \iff a=b^2+1$$

Answer 4

Such points $(b,a)$ lies on parabolic curve $y=x^2 +1$

Answer 5

In the rationals, $b$ can be anything and $a$ must be a perfect square plus one.

In the reals, the only requirement is $a\ge1$.

Answer 6

$a$ and $b$ are related by a single real variable $u$ that allows to relate to other numbers

$$ a= \cosh^2 u,\, b= \sinh u .$$