Find all positive integers n such that 12n-119 and 75n-539 are both perfect squares. Let N be the sum of all possible values of n. Find N.

Let 75n - 539 = l^2 and 12n - 119 = k^2 . where n is a natural number.

Multiply 75n - 539 = l^2 by 4 to give 300n - 2156 =4l^2 and 12n - 119 = k^2 by 25 to give 300n - 2975 = 25k^2.

Subtract the two new expressions to give 4l^2 - 25k^2 = 819 which can be factorised (using the difference of two squares) to give (2l - 5k)(2l + 5k) = 819.

The prime factorisation of 819 is 3^2 * 7 * 13 There are five cases to consider. Dealing with the cases (noting that 2l - 5k < 2l +5k ) yields that n can only be 20 or 12.

Hence N = 20 + 12 = 32.

SK

Related STEP University answers

All answers ▸

Suppose that 3=2/x(1)=x(1)+(2/x(2))=x(2)+(2/x(3))=x(3)+(2/x(4))+...Guess an expression, in terms of n, for x(n). Then, by induction or otherwise, prove the correctness of your guess.


How would you prove the 'integration by parts' rule?


Find h(x), for x≠0, x≠1, given that: h(x)+h(1/(1−x))=1−x−1/(1−x)


Prove that any number of the form pq, where p and q are prime numbers greater than 2, can be written as the difference of two squares in exactly two distinct ways.