Natural Number Solution in Polynomial=Polynomial Equation?

[StudyOriented]

Hello,

First of all, sorry if my English is bad, that because i am not from an English-speaking country.
But, i think you could understand what i say, because mathematics is universal.

Help me, my teacher gave us a bonus quiz, about a loan fine problem.

Here is the problem

A bank have a rule that apply for every loan

A person borrow $x loan from the bank.
If he does not pay the loan at the deadline.
Then he should pay for the fine.

The rule for the fine;
1st Month = $ x
2nd Month = $ x^2 + 1st Month fine
3rd Month = $ x^3 + 2nd Month fine
4th Month = $ x^4 + 3rd Month fine
...
and so on, the rule continues...

so, if a person borror $2 loan and late for 5 Months he should pay the fine for $(2^5)+$(2^4)+$(2^3)+$(2^2)+$2 = $62 , the fine rule resembles a polynomial function.

Now, here is the quiz from my teacher

Andy borrow $ A from the bank, and Bella borrow $ B from the bank. (A & B are natural numbers)

If Andy was late to pay the loan for 4 Months and Bella was late to pay the loan for 2 Months.
Then the fine Andy and Bella should pay was the same.

The question is,
How much the money that Andy and Bella borrow each and how many possibilities for the answers?

(How many is A and B and how many possibilities of natural numbers to apply for A & B? d

that is the quiz, my teacher also gives a clue, "Discriminant" he said

I wrote the equation for the fine;

(A^4)+(A^3)+(A^2)+A = (B^2)+ B​

...
But what are the solution(natural numbers) for A and B?

When i try finding the solution by trial and error i found one possibilities for A & B, that A=2 and B=5.


But how can i find the solution mathematically?

How many possibilities of natural numbers can apply for the solution?

Is there any pattern for this kind of problem? (finding natural number solution in polynomial equation)

Does the Discriminant play an important role? (my teacher said that discriminant is the clue)

Please Help, Thank you ​

 

[StudyOriented]

You have: b + b^2 = a + a^2 + a^3 + a^4

Formula: a^1 + a^2 +....+ a^n = [a^(n+1) - a] / (a - 1)

In your case, n = 4 (given); so:
b^2 + b = (a^5 - a) / (a - 1)

Let k = (a^5 - a) / (a - 1)

then:
b^2 + b - k = 0 : a quadratic; using quadratic formula:
2b = -1 + SQRT(1 + 4k) : the "-" solution does not apply, since answer positive

The discriminant 1 + 4k must be a square (to get natural numbers).
Possibilities are 9 if k=2, 25 if k=6, 49 if k=12, 81 if k=20, 121 if k=30, and so on...

You'll find that k=30 is the only possibility, resulting in a=2 and b=5.
Sorry, can't find a clean-cut easy way...

Thank you very much!!!
i appreciate it a lot

even i ask my friend who can make a c++ program for this, in 1000 number checked she just found 1 possibilities, 2 and 5 :shock:

BTW , how did you find after the possibilities? Did i should try each number to k to create square?