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.........

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## Comments

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TopNewestYou 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

positiveinteger. I was thinking that \(N=101\) works.Log in to reply

Seriously could not follow you. Please extrapolate a bit.

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

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yes

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

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

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it came in isi right

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

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

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age) ? Anyway, where from did you get the questions ?Log in to reply

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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.

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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.

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

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