Find the largest integer N such that N + 496 and N + 224 are perfect squares.?

Answer 1

N + 496 = X²
N + 224 = Y²

X² - Y² = 496-224 = 272

( X + Y ) ( X - Y ) = 272

272 = 2^4 * 17

Possible pairs of products to generate 272 are

( 1, 272 ) with X, Y = no solution
( 2, 136 ) with X, Y = 69, 67
( 4, 68 ) with X, Y = 36, 32
( 8, 34 ) with X, Y = 21, 13
( 16, 17 ) with X, Y = no solution

Looks like we get the largest N from 69, 67

Substituting back into our initial expressions...

N + 496 = X² = 69² yields N = 4265
N + 224 = Y² = 67² yields N = 4265

later: RobertJ may have had first response, but I had the right answer up 'way before he did :)

Answer 2

N+496 = L^2, L > 22
N+224 = M^2, M > 14
Subtract the second equation from the first,
272 = (L+M)(L-M)

To get the largest N, factor 272 into two even integers as far from each other as possible. The reason is that L+M and L-M must be both even integers, and the smaller the difference, the larger the sum.

2*136 = (L+M)(L-M)

Therefore,
L+M = 136......(1)
L-M = 2......(2)

(1)+(2): 2L = 138
L = 69, M = 67

So, N = 67^2 - 224 = 4265