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I'm having issues with this problem.

$N + 2000$ is an square number. $N - 17$ is also an square number. Find $N$.

2 Answers 2

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Hint:

Let,

$$N+2000=a^2$$

$$N-17=b^2$$

So that,

$$(N+2000)-(N-17)=a^2-b^2$$

$$2017=(a-b)(a+b)$$

$2017$ is prime so there isn't really that many ways to break it apart into multiplication of integers.

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    Forgot to add, n must be an positive integer – 2017-02-17
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    Yes I know, this hint should be enough @HectorMagaña – 2017-02-17
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    Snap, I think I got it, will confirm later since im doing some stuff right now – 2017-02-17
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    $1016081$ is the right answer. $1016064$ is $1008^2$. $1018081$ is $1009^2$ – 2017-02-17
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    Note that $1008+1009=2017$ Is it a contest problem ? :D – 2017-02-17
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$N+2000=p^{2}$ (1)

$N-17=q^{2}$ (2)

Substract (2) from (1)

$N+2000-(N-17)=p^{2}-q^{2}$

$2017=(p-q)(p+q)$

2017 is a prime, thus:

$p-q=1$

$p+q=2017$

Solving the system of equations gives you p=1009, q=1008. Now use p in (1), or q in (2) $N+2000=1009^{2}$ And the desired number is 1016081.