8. Let K be an integer that is greater than 1, a perfect square, and equalto E-1 i for some integer D.a) Find the first two integers K and D that satisfy these conditions andthenb) Evaluate the expression A in both cases.A= [1+2+ ... + (VK – 1)] – [(VK +1) + (VK +2) +... + D]c) Can you make a conjecture about the value of A? Does it depend onthe values of K and D?

A perfect square is a positive integer whose square root is a positive integer. Example of perfect squares is 4, 9, 16, etc. It is not necessary that a perfect square must be a sum of natural numbers starting from 1. But, there are some perfect squares that can be expressed as the sum of certain natural numbers starting from 1. Let K>1 be an integer that is a perfect square. Also, it is given that K=∑i=1DiThe given conditions impliy that K is the sum of first D natural numbers. Now, ∑i=1Di=DD+12 By inspection method, we can see that the first value of D for which DD+12 is a perfect square is 8. Also, for D=8, 88+12=36 Thus, the first two integers satisfying the given conditions are K=36, D==8The expression is A=1+2+...+k-1-k+1+...+D Substituting the value of K and D, A=1+2+...+k-1-k+1+...+D=1+2+..+36-1-36+1+...+8=1+2+...+5-7+8=5·62-15=0 Thus, for the first two integers K and D satisfying the given conditions, the value of A is 0. To make a general conjecture, let us assume K=a and D=a+r, r≥0. Then, A=1+2+...+k-1-k+1+...+D=1+2+...+a-1-a+1+a+2+...+a+r=aa-12-ra+rr+12 So, for any perfect square K=a2 satisfying the condition K=∑i=1Di with D=a+r, r≥0, A=aa-12-ra+rr+12 It can be inferred that A depends on K as well Dbecause in the above conjecture, if we know K, then we can be able to find K=a and for determined D, we can express D as D=a+r which helps us to find r.