Number Theory problem

For how many positive integral values of N is the expression N(N-101) the perfect square of a positive integer ?

I got 2601 as the only value. I want to know if there are other such values ? I know it's a silly question for Brilliant problem solvers. Please help.........

#NumberTheory #HelpMe!

Easy Math Editor

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Comments

You are missing another obvious (positive integral) solution. (This statement is false. See edit below.)

Hint: Greatest Common Divisor.

Edit: I missed out that $N (N-101)$ has to be the square of a positive integer. I was thinking that $N=101$ works.

Calvin Lin Staff - 8 years ago

Seriously could not follow you. Please extrapolate a bit.

Nishant Sharma - 8 years ago

I don't think that any other solution exists, Calvin Sir. Both $N$ and $N(N-101)$ are to be positive.

Shourya Pandey - 8 years ago

yes

superman son - 8 years ago

Me too. That was why I posted this since I had done few problems only. BTW did you also write ISI 2013 ?

Nishant Sharma - 8 years ago

I misread, didn't see "positive" in the latter instance. Thanks for correcting.

Calvin Lin Staff - 8 years ago

it came in isi right

superman son - 8 years ago

yea,you too gave isi,how many could you answer in part i & II...?

Sayan Chaudhuri - 8 years ago

Yeah it was an ISI problem. Well I attempted 29/30 in part I and solved 4 in part II. How about you both ?

Nishant Sharma - 8 years ago

@Nishant Sharma – i am 13 so i cannot participate but i saw the question and tried to solve it . i got about 28 in part 1 and 7 in part 2

superman son - 8 years ago

@Superman Son – Are you sure or just kidding(your age) ? Anyway, where from did you get the questions ?

Nishant Sharma - 8 years ago

@Nishant Sharma – y kiddin and q from http://cheentaganitkendra.blogspot.in/2013/05/isi-2013-bmath-bstat-subjective-paper.html

superman son - 8 years ago

Let $x^2=N$ and

$y^2=N-101$ where $x>0,y>0$

Now we can write from our first statement that :

$y^2=x^2-101$

$\Rightarrow x^2-y^2=101$

$(x+y)(x-y)=101$

Therefore it follows that $(x+y)$ and $(x-y)$ are factors of $101$ .

Now since $101$ is a prime and $x,y$ are positive integers then the only possible values that $(x+y)$ can take is $101$ .

Note that since $x>y$ $,$ $(x-y) \neq 101$ .

So the only solution possible is when $x=51$ which gives $y=50$ .

Therefore $N=x^2=2601$ is the only solution.

Aditya Parson - 8 years ago

You need to substantiate your first line. Why must we have $N=x^2$ ? You are merely given that $N$ is an integer, and not told that it is a square. For example, if we ignore the condition that $N(N-101)$ must be positive, then $N=101$ will be a solution, but that disagrees with your claim that $N$ must be a perfect square.

Calvin Lin Staff - 8 years ago

If $N=101$ is a solution then $y=0$ but I already stated that $y>0$ . So N=101 cannot be a solution.

Aditya Parson - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$