Prime factorization of a large integer. Prove $A − \sqrt{N} < 1$ , given $N = pq$, $A =\frac{p+q}{2}$ and $|p − q| < 2 \sqrt[4]{N}$.

Question

Prove $A >\sqrt{N}$ and $A − \sqrt{N} < 1$, given that $|p − q| < 2 \sqrt[4]{N}$ and $A =\frac{p+q}{2}$,
where $p$ and $q$ are primes, N is a large integer and $N = pq$.

By $(p+q)^{2} > 4pq$, when $p \neq q,$ I can prove $\frac{p+q}{2} > \sqrt{pq}$, which indicates $A >\sqrt{N}$.
However, I have no clue about how to prove $A − \sqrt{N} < 1$.

I have searched relevant theories about prime factorization of large integers online, but still, I couldn't find the answer that I need, perhaps for the question is too specific.
If you have any idea, please post it here and I will really appreciate it.

Answer 1

From the binomial formulas we have $$(p-q)^2=(p+q)^2-4pq=(2A)^2-4N$$ $$A^2-N=(A+\sqrt{N})(A-\sqrt{N})=\frac{1}{4}(p-q)^2$$ $$A-\sqrt{N}=\frac{1}{4}(p-q)^2/(A+\sqrt{N})$$ Now use $A>\sqrt{N},$ i.e. $A+\sqrt{N} >2 \sqrt{N}$ to decrease the denominator and the condition for $p,q$ written as $(p-q)^2 < 4\sqrt{N}$ to increase the numerator to get $$A-\sqrt{N}<\frac{1}{4}(p-q)^2/(2\sqrt{N})=\frac{1}{4}\times 4\sqrt{N}/(2\sqrt{N}) = \frac{1}{2} < 1 $$