Question

If 200 is added to a positive integer I, the result is a square number. If 276 is added to to the same integer I, another square number is obtained. Find I.

Answer 1

`\begin{gathered} Let\text{ } \\ 200\text{ + I= x}^2----------\left(1\right) \\ 276+I\text{ =y}^2----------\left(11\right) \\ Subtract\text{ equation \lparen1\rparen from equation \lparen11\rparen} \\ 276+1-\left(200_+I\right?=y^2-x^2 \\ 76=\left(y-x\right?\left(y+x\right? \end{gathered}`

Now y+x and y-x differ in 2x.

One of them is even, because their product is even, so the other must be even too.

76=2*2*19 and 19 is prime.

We can assume x,y>=0,

Thus, y+x=2.19, and y-x=2

from here y=20, x=18

Therefore,

`\begin{gathered} 200+1=18^2 \\ 200+I=324 \\ I=324-200 \\ I=124 \end{gathered}`

The answer is I = 124