1. ## Perfect squares

Determine all positive integers n such that n+100 and n+168 are both perfect squares:
Here' my attempt:
n+168 should be less than 35sqr
Since, 35sqr - 34sqr = 69 which is greater than 68, the difference between n+100 and n+168

Beyond that, I've got no clue.

2. Originally Posted by darknight
Determine all positive integers n such that n+100 and n+168 are both perfect squares:
Here' my attempt:
n+168 should be less than 35sqr
Since, 35sqr - 34sqr = 69 which is greater than 68, the difference between n+100 and n+168

Beyond that, I've got no clue.
Hi darknight,

I would proceed by finding all qualifying perfect squares, then from them finding n.

$\displaystyle n+100=a^2$

$\displaystyle n+168=b^2$

So

$\displaystyle a^2-100=b^2-168$

$\displaystyle b^2-a^2=68$

$\displaystyle (b-a)(b+a)=68$

We know that $\displaystyle (b-a)$ and $\displaystyle (b+a)$ are integers, so this problem is now equivalent to factoring 68 into two divisors, $\displaystyle 68=d_1d_2$.

For any pair $\displaystyle (d_1,d_2)$, we can solve

$\displaystyle (b-a) = d_1$

$\displaystyle (b+a) = d_2$

$\displaystyle \Rightarrow 2b = d_1+d_2$

First, try $\displaystyle 68=1\cdot68$

$\displaystyle \Rightarrow 2b = 69$

No integer solutions.

Next try $\displaystyle 68=2\cdot34$

$\displaystyle \Rightarrow 2b = 36$

$\displaystyle b = 18$

$\displaystyle a = 16$

$\displaystyle n = a^2 - 100 = 156$

Next try $\displaystyle 68=4\cdot17$

Again there are no integer solutions.

So the only possible n is $\displaystyle \boxed{156}$.

3. Ah well, went till (b-a)(b+a) = 68 and then hit the brick wall..
Thanks

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### Determine all positive integers of 'n' such that n 100 and n 168 are both perfect squares.

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