Consider the family of circles : $x^2+y^2-3x-4y-c_1=0$, $c_1\in N(i=1,2,3,\dots n)$ 

Also, let all circles intersects x-axis at integral points only and $c_1<c_2<c_3<c_4\dots <c_n$ . A point (x,y) is said tobe integral point, if both coordinates x and y are integers.

Solution:

Step 1: Rewriting the Circle Equation

First, we rewrite the circle equation in a more standard form by completing the square: $$x^2 - 3x + y^2 - 4y = c_1$$ Completing the square for $x$ and $y$: $$(x - \frac{3}{2})^2 - \frac{9}{4} + (y - 2)^2 - 4 = c_1$$ $$(x - \frac{3}{2})^2 + (y - 2)^2 = c_1 + \frac{13}{4}$$

Step 2: Conditions for Circle Intersection with the X-Axis

A circle intersects the x-axis where $y = 0$. Substituting $y = 0$ into the modified circle equation gives: $$(x - \frac{3}{2})^2 + (2)^2 = c_1 + \frac{13}{4}$$ $$(x - \frac{3}{2})^2 = c_1 + \frac{13}{4} - 4$$ $$(x - \frac{3}{2})^2 = c_1 - \frac{3}{4}$$ For the circle to intersect the x-axis at integral points, the expression $(x - \frac{3}{2})^2$ must be such that the solutions for $x$ are integers. This implies that $(x - \frac{3}{2})^2$ itself must be a perfect square.

Step 3: Ensuring Integral Intersections

The left side, $(x - \frac{3}{2})^2$, can be written as $(2x - 3)^2 / 4$, a transformation of the variable $x$ into an expression involving integers only when multiplied by 4. So, for integer solutions of $x$: $$(2x - 3)^2 = 4(c_1 - \frac{3}{4})$$ $$(2x - 3)^2 = 4c_1 - 3$$ The right side must also be a perfect square for $x$ to be integral. Let's say $4c_1 - 3 = k^2$ for some integer $k$. Solving for $c_1$ gives: $$4c_1 = k^2 + 3$$ $$c_1 = \frac{k^2 + 3}{4}$$ For $c_1$ to be a natural number, $k^2 + 3$ must be divisible by 4. Analyzing this condition modulo 4, we observe that $k^2 \mod 4$ can be either 0 or 1 (since $k^2$ is the square of either an even or an odd number). The only way $k^2 + 3$ is divisible by 4 is if $k^2 \mod 4 = 1$. Therefore, $k$ must be odd.