Why n?

Number Theory Level 3

For integer $0 \leq n \leq 200$ , find the sum of all $n$ such that $n(n-101)$ is a perfect square.

Thanks to Swapnil D. for introducing me to this problem.

The answer is 101.00.

2 solutions

Mehul Arora
Aug 30, 2015

METHOD 1

$n(n-101)=m^2$

$n^2-m^2=101n$

$(n+m)(n-m)=101n$

$\because 101$ is prime,

Case 1 : $(n+m)=101 \\ (n-m)=n$

If $(n-m)=n, \therefore m=0$

Case 2 : $(n-m)=101 \\ (n+m)=n$

$If (n+m)=n, \therefore m=0$

Thus, we conclude that $n(n-101)=0$

$\therefore n=0,101$

Adding both, we get 101.

Moderator note:

Both of these methods are incorrect.

Method 2 is wrong. Do you see why? Hint: What assumption about the number of roots are you making?

Method 1 is slightly wrong. Do you see why?
Hint: Uniqueness of prime factorization only works if
1. All of the terms are primes
2. All of the terms are non-zero (or positive)

In your method 1, how do you prove that you can't possibly break $n$ into other factors. You only included two cases, where the factors were $101, n$ , but never accounted for the possibility that $n$ could be broken apart.

Alan Yan - 5 years, 9 months ago

That is a correct criticism of method 1. Thanks for pointing it out!

Calvin Lin Staff - 5 years, 9 months ago

Are you sure you want n to be a real number? If so, this problem is completely wrong.

Method 2 is wrong. Do you see why? Hint: What assumption are you making?

Method 1 is slightly wrong. Do you see why?
Hint: Uniqueness of prime factorization only works if
1. All of the terms are primes
2. All of the terms are non-zero (or positive)

Calvin Lin Staff - 5 years, 9 months ago

Method 2 is incorrect. You cannot use Vieta's for a polynomial with 2 variables since $m^2$ is not a constant. (Correct me if I am wrong)

Nihar Mahajan - 5 years, 9 months ago

You are somewhat incorrect. Vieta's formula would still hold, with the conclusion being that there would be a pair of roots which sum up to 101.

The issue is that we don't know how many pairs of roots there are. If there are 2 pairs (and assuming that all of these fall within the given range, which is not guaranteed), then the answer would have been 202. For example, in the previous version where real numbers were allowed, then number of the form $\frac{ 101 \pm \sqrt{10205 }} { 2}$ would have been acceptable solutions.

Calvin Lin Staff - 5 years, 9 months ago

Yeah the answers would have been 202 because the possible values of n are 101,0,2601,-2500.

Kushagra Sahni - 5 years, 9 months ago

Then how about n = 2601? Doesn't it gives:

$n(n-101)=2550^2$

Gian Sanjaya - 5 years, 9 months ago

Yeah , you are right.

Nihar Mahajan - 5 years, 9 months ago

If we have the n without criteria do we assume $n \in Z$ ? Or still use $n \in R$ ?

Gian Sanjaya - 5 years, 9 months ago

Thanks :) I have edited the question.

Mehul Arora - 5 years, 9 months ago

Cheers! @Mehul Arora

Swapnil Das - 5 years, 9 months ago

Cheers! @Swapnil Das

Mehul Arora - 5 years, 9 months ago

Aditya Gupta
Dec 4, 2018

Let n(n-101)=m^2, m being positive. Now simply apply quadratic formula for n, and discriminant would be 101^2+4m^2, which should be a perfect square say p^2, p being positive. So (p-2m)(p+2m)=101^2. As 101 is a prime number, so either p-2m=1 and p+2m=101^2 or both equal to 101. From here it is easy to see that n can only be 101 and zero in the range 0 to 200

Why n?

Thanks to Swapnil D. for introducing me to this problem.

The answer is 101.00.

2 solutions

Moderator note:

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