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Number Theory Level 3

Find a positive integer $n$ such that both

$n + 88 \text{ and } n - 28$

are perfect squares.

3 solutions

Sunil Pradhan
Nov 3, 2014

N + 88 = x² and N – 28 = y²

subtract (x ² – y²) = (x + y)(x – y) = 116 = 116 × 1 or 58 × 2 or 29 × 4

then Let (x + y)(x – y) = 116 × 1

x + y = 116 and x – y = 1 solving for x and y both are not integer or x + y = 29 and (x – y) = 4 solving for x and y both are not integer

When x + y = 58 and (x – y) = 2 solving for x and y both are integer

x = 30 and y = 28

N + 88 = 900 so N = 812

and 812 – 28 = 28²

Christian Daang
Oct 26, 2014

Let:

a^2 = n+88

b^2 = n-28

Subtracting the 2 equations,

(a^2 - b^2) = 88+28 = 116

(a+b)(a-b) = 116(1) or (58)2 or (29)4

If you will check for them, 58 and 2 are just the #'s compatible.

So...

(a+b)(a-b) = 58(2)

a+b = 58

a-b = 2

a = 30 and b = 28

Therefore, the number is just: 30^2 - 88 or 28^2 + 28 which are bpth equal to 812...

Therefore, the final answer is 812.

@Christian Daang The submitted dispute stated "it is not clear what you want us to do with all of the answers. Do you want the sum?"

I agree that the statement was not clear, and have edited your problem to account for this.

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Calvin Lin Staff - 6 years, 7 months ago

Shohag Hossen
Dec 10, 2014

30 * 30 = n + 88 = 900

28 * 28 = n - 28 = 784

n + 88 = 900

n = 812 . (answer)