It could be anything!

Find the largest integer n n satisfying the following conditions :

  1. n 2 n^2 can be expressed as the difference of two consecutive cubes.

  2. 2 n + 79 2n+79 is a perfect square .

141 217 181 271 325

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1 solution

Department 8
Mar 7, 2016

This is only a solution which uses half-solving.

n 2 = ( x + 1 ) 3 x 3 n 2 = 3 x 2 + 3 x + 1 3 x 2 + 3 x ( n 2 1 ) = 0 n^2=(x+1)^3-x^3 \\ n^2=3x^2+3x+1 \\ 3x^2+3x-(n^2-1)=0

Now for n n to integer the discriminant if the above quadratic equation should be a perfect square. Now start keeping the values of n n to find which makes 12 n 2 3 \sqrt{12n^2-3} an integer.

Yeah and 2 options are eliminated because 2n+79 isn't a perfect square. Very Counter-intuitive, at least the options should have been good.

Kushagra Sahni - 5 years, 3 months ago

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