I tackled the first problem and I noticed that I could try and make the quadratic equation at the bottom to be equal to 1. Because, 1 times the power of any integer number is always going to be 1. So I made an equation like this (by the way n^2 is meant to be n squared):
n^2-5n+5=1
Next, I tried looking for possible values for n to be equal by 1, by taking 1 off each side:
n^2-5n+4=0
I realised you can factorise the equation on the left side, so I did it to get to the final step.
(n-1)(n-4)=0
From this equation, we can deduce that n=1 and n=4. So we have two solutions to the first problem. I was really excited! Next, I tried looking for ways for making the power of the integer at the bottom to be equal to 0. Because a positive number with a power of 0 will always be equal to 1. So I made another algebraic equation.
n^2-11n+30=0
You can factorise this like the other equation to get this simplified equation:
(n-6)(n-5)=0
From this, n=6 and n=5. Now we have four solutions (Yay!), 1,4,5 and 6. I realised we had two more possible values left and I thought it couldn't be a negative number as it would result in a really high number with a high power. It couldn't be a fraction as it wouldn't be equal to a integer. I thought that maybe we could have something like -1 to the power of an positive even integer.Then it suddenly popped in my head that the two numbers I'm missing are 2 and 3. Because that would mean all the 6 values of n are in number order from 1 to 6. I might be onto something!
I tried substituting 2 as a value of n into the mega quadratic equation and I came up with -1 to the power of 12 which would come up with 1! I did the same thing with 3 and I got -1 to the power of 6! I had all 6 solutions for the first problem!
For the second question, I did the same thing and made equations for the bottom number to be equal to 1 and another one with the power equal to 0 and they appear respectively.
(n-2)(n-5)=0
(n-6)(n-7)=0
From these equations, n can be equal to 2,5,6 and 7. I noticed I was missing two numbers like last time and I could fill in a number sequence with 3 and 4. I got -1 to the power of 12 substituting 3 as n and -1 to the power of 6 substituting 4 which both equal to 1. Now I have found the answers for the second question!
I decided instead of looking for other mega quadratic equations, I would make an equation to find an infinite amount of mega quadratic equations. I looked at how I found the values of n for both equations to find a pattern. I noticed that when I was making equations for the bottom number to be equal to 1, The values I found to be n in the first question was 1 and 4. The values in the second question was 2 and 5. The values of n were added once between the questions!
I did the same thing with the power quadratic equation and found in the first question and second question, I got 5,6 and 6,7 respectively. I tried finding relationships with the values of n with the mega quadratic equation and I found out that I could use the values of n that I found by making the bottom number 1 to make the numbers in the mega quadratic equation. Eg. The two fives in the bottom bit of the first mega quadratic equation.
(n^2−5n+5)
The values of n I found in the first question (1 and 4) make the numbers as they add up to make the 5 in -5n (the minus will always stay there not affected by adding the two values of n) and to get the 5 on the right you have to multiply the two values of n together and add 1. The reason you don't add them both together, is because in the second question I found out that 2+5 isn't going to make the 11 on the power equation.
(n^2−7n+11)
But, 2 multiplied by 5 and adding one to the answer will get you the number on the right side of the quadratic equation on the bottom. This can be done for the first equation. I did something similar to the power quadratic equation and found that to get the number in the middle, you have to add the values of n you find while using the equation which makes the power quadratic equation equal to 0.
For the power equation, you can multiply the two values of n together to get the number on the right side of the equation (you don't have to add 1). From what I found, I made an equation to make a mega quadratic equation (I think I made an ultimate quadratic equation!).
(n^2 - (2x+3)n+(x(x+3)+1) to the power of (n^2 - (2x+9)n + (x+4)(x+5))=1
(x is an integer and the lowest possible value of n that will make the whole equation equal to 1)
Eg. The first question's values of n that makes it equal to 1, is 1,2,3,4,5 and 6. 1 is the lowest value of n that makes the equation equal to 1.
Using this equation, you can find an infinite amount of mega quadratic equations like:
(n^2−9n+19) to the power of (n^2−15n+36)=1
(n^2-11n+29) to the power of (n^2-17n+72)=1
You can find many mega quadratic equations and its answers by using the equation!