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Can anyone help on this?

What is the positive integer such that the sum of the positive integer and 100 is a square number, and the sum of the positive integer and 168 is also a square number?

Here is what I did: assume the positive integer being $x$, then $$ x+100=y^2$$ $$x+168=z^2$$ This gives $(z-y)(z+y)=68=17*2^2$. There are many possibilities. Is there an simple way to find the answer from here?

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  • $\begingroup$ Not that many possibilities: $68$ has $6$ divisors only, so only $3$ pairs to consider. $\endgroup$ – lhf Sep 19 '16 at 18:30
  • $\begingroup$ Hint: $z \pm y$ have the same parity. $\endgroup$ – dxiv Sep 19 '16 at 18:30
  • $\begingroup$ What does this have to do with prime numbers? $\endgroup$ – lulu Sep 19 '16 at 18:33
  • $\begingroup$ Your title says positive prime number. If you can, please change your title to describe the actual question you have. Or, if you are really asking for a prime integer, add that to the body of your question. $\endgroup$ – 6005 Sep 19 '16 at 18:39
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Not that many possibilities: $68$ has $6$ divisors only, so only $3$ pairs to consider:

$z-y=1, z+y=68$

$z-y=2, z+y=34$

$z-y=4, z+y=17$

Add @dxiv's remark and only one pair is left.

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