Need help solving diophantine inequalities (sqrt(a) < b < sqrt(a + 10^8))

[Komwom]

I have the diophantine inequality

sqrt(a) < b < sqrt(a + 10^8)

where a and b are non-negative integers. How do I find out for which a this inequality has at least one solution for b? I want to use mathematical reasoning rather than testing.

 

[blamocur]

A hint: if there exists at least one solution what is the minimal value for $\sqrt{a+10^8} - \sqrt{a}$ ?

 

[blamocur]

I have several questions:

1. Do you need describe all such $a$'s ?
2. Is this problem from your homework ?
3. Could you post the full statement of the problem?

Thank you.

 

[blamocur]

Another, simpler, approach: pick an arbitrary $b$ and describe the set $A_b$ of all $a$'s which satisfy the inequality. Then figure out which $a$'s don't belong to any $A_b$.

P.S. Is there a penalty for having more posts in the thread than the OP?

 

[khansaheb]

P.S. Is there a penalty for having more posts in the thread than the OP?

Yep - the corner

Come on in and join us......

 

[JeffM]

I am far from sure I understand what the real question is because it is obvious that you can never answer a question about an infinitie number of pairs of numbers by testing every pair. However, the question seems to be to determine some sort of description of or limits on n(b) where n(b) is the number of possible values of a such that $\sqrt{a} < b < \sqrt{a + 10^8}$, given that a and b are both non-negative integers.

$$0 \le \sqrt{a} < b \implies 0 \le a < b^2.\\ b = 0 \implies b^2 = 0 \implies a < 0, \text { which is a contradiction.}\\ \therefore \ n(0) = 0.\\ b \ge 1 \implies b^2 \ge 1 \implies 0 \le a < b^2 \implies n(b) \le b^2 - 1.\\$$
I admit that I have not worked this out, but I believe that working with the other constraint, namely

$$0 < b^2 < a + 10^8 \implies -10^8 < (b - 10^4)(b + 10^4) < a \ge 0$$,

may also lead to bounds on n(b). Let us know how far you get that way.