Question: Factorization of N=pq are partially in P?

I found a condition for p, q that N=pq can be factored in plynominal time using Maple 2020.
Is fllowing Hypothesis and Proof is right?

Hypothesis

       N=pq  p and q are large prime respectively.
         R=q/p  q > p  R is very close to an small integer or a simple rational number.
       
        N=pq can be factorized in time polynomial

Proof
        point[p, q] is on y=N/x
        y=N/x  and y=Rx cross at point[p, q]
        N is n digit
        upper  2 digits N2  round off the 3rd digit
        upper  3 digits N3  round off the 4th digit
        upper  4 digits N4  round off the 5th digit
        
        y=N2/x and y=Rx cross at point[p2,q2]
        y=N3/x and y=Rx cross at point[p3,q3]
        y=N4/x and y=Rx cross at point[p4, p4]

        But we only know N.

        Let line up candidates point[p2,q2] , point[p3,q3] and point[p4, p4]

       N2 < 99  i=1..10 j=1..10
       R2=i/j
       f2=N2/R2 - j^2
       dn2=abs(N2-R2*j^2)   

      N4 < 9999  i=1..99 j=1.. sqrt(N4)
      R4=i/j
      f4=N4/R4 - j^2
      dn4=abs(N4-R4*j^2)

     Point[j, i] that have  small f2 and dn2 can be nominated as candidate for point[p2, q2]
     Point[j, i] that have  small f3 and dn3 can be nominated as candidate for point[p3, q3]
     Point[j, i] that have  small f4 and dn4 can be nominated as candidate for point[p4, q4]

    Find cross point[px, qx] of y=R2x and y=N/x , y=R3x, y=N/x and y=R4x, y=N/x
    Find the nearest prime pn for px and the nearest prime qn for qx
   
   pn*qn=N  bingo!

   Number of candidates are finit.
   You can factorized N=pq in time polynomial.
                                   
                                                                       Q.E.D. ?

In addition, using "https://www.mapleprimes.com/questions/228532-Strange-Factorization"

Rang from p - half digits of p to p + half degits of p and /or range q - half digits of q to q + half degits of q  N=pq can be factored in plynominal time.