In the given equation, the total number of "a" that exists is n.
Therefore the equation can be simplified to: \(an + (1+2+...+(n-1))\)
1 + 2 + 3 + ... + n = \(\dfrac{n(n+1)}{2}\)
Therefore the whole thing after the plus sign simplifies to: \(\dfrac{(n-1)(1+(n-1))}{2}=\dfrac{n(n-1)}{2}\)
Original equation becomes \(an+\dfrac{n(n-1)}{2}=100\)
And then further simplify: \(\dfrac{1}{2}n^2+an-\dfrac{1}{2}n-100 = 0\)<--- I am trying to make it a quadratic here.
Take out the common factor n: \(\dfrac{1}{2}n^2+n(a-\dfrac{1}{2})-100 = 0\)
To make it easier to solve, multiply each term by 2.
\(n^2 + n(2a - 1)-200=0\)
It now becomes a quadratic equation with a is a constant and solving for n. Then we can find n in terms of a.
\(n=\dfrac{-(2a-1)\pm\sqrt{(2a-1)^2-(4)(1)(-200)}}{2}=\dfrac{1-2a\pm\sqrt{4a^2-4a+801}}{2}\)
Because n is an integer, \(\sqrt{4a^2-4a+801}\) must be an integer too, to make n an integer.
That implies that \(4a^2 - 4a + 801\) is a square number.
Also, \((4a^2 - 4a + 801)\pmod4\equiv 1\).
We now think of a square number which mod 4 = 1.
The nearest square number to 801 which mod 4 = 1 is 841, therefore we assume that the quadratic there is = 841 first.
\(4a^2 - 4a + 801 = 841\\ a^2 - a = 10\\ a(a-1)= 10\)
No positive integer "a" satisfies the equation.
The second nearest square number to 801 which mod 4 = 1 is 729,
\(4a^2 - 4a + 801 = 729\\ a^2 - a = -72\\ a(a-1)= -72\)
Still no positive integer "a" satisfies the equation.
We should think of a bigger number.
31^2 = 961 and mod 4 = 1.
\(4a^2 - 4a + 801 = 961\\ a^2 - a = 40\\ a(a-1)= 40\)
Still no positive integer "a" satisfies the equation :(
Try 1089
\(4a^2 - 4a + 801 = 1089\\ a^2 - a = 72\\ a(a-1)= 72\\ a = 9\)
Yay! :D
We back-substitute a = 9 into the quadratic formula to find the corresponding n.
\(n=\dfrac{1-2(9)\pm33}{2}\\ n = 8\text{ or }n=-25(\text{rejected})\\ n = 8\)
We get: a = 9, n = 8 (Finally after a lot of work XD)
