Never tear a page

I was writing a solution to the following answer before but was rejected by the challenge master there as "like hit and trial". I therefore would like to try my solution with other masters

Let $x$ be the number of all pages, and $y$ be the page number torn apart. It will come to sense that $y < x$ . By writing the equation from it, it will be: $1+2+3+\dots+y+y+1+\dots+x = 15,000+y+y+1$ or, by using the sum of consecutive integer and equation rewrites, would be $x(x+1) = 30,000+4y+2$ . Given that $y > 0$ , then $x(x+1)-30000 > 0$ This equation is important since we will try to estimate the page number from it. When factorizing, it will come to light that the answer would be x must be more than $0.5(1+\sqrt{120001})$ , or simply put $\sim 0.5(1+346.411) \sim 173.705$ , which means the page number must be at least 173 pages

By trying to find the number that satisfy $y < x$ , $x \geq 173$ and $x(x+1) = 30,002+4y$ , it will hold that $(x,y) = (173, 25), (174,112)$ , but $(174, 112)$ is not possible as the page 112 would be with the page 111, and page 113 should be with the page 114, which means 112 and 113 cannot be tear off together. Hence, there is only one answer, the book has 173 pages and pages 25 and 26 are being torn apart

what we can do is that . . . . first find the page numbers of the respective pages of whose sum (page number's) is given. i.e. for number 2459 the page numbers are - 2459/2=1229.5 because page numbers must be consecutive & integral then natural answers for 2459 are 1229 & 1230 [1229+1230=2459]. similarly for 235 - 117 &118 . . . .(1) 137 - 68 & 69. . . . . . .(2) 51 - 25 & 26 . . . . . . .(3) Now , for 1229 & 1230 minimum number of page = 1230. Similarly for (1) (2) & (3). 1229 & 1230 are not correct because then the sum of page numbers =1228+1227+1226 +. . .+ 1000 . . .[ clearly > 15000 ] Then for the reaming 3 I've used Microsoft Excel.