first time asking a question here.

This proof seems simple, but the only part throwing me off is the the first two remarks "Show that for every positive integer a, there exist a positive integer $b$ such that $ab+1$ is a perfect square."

What I have is: Let $k = n^2$ where is an integer and $n^2$ is perfect square. then $ab+ 1 = k $

This is where I get stuck.

  • 4
    $\begingroup$ Hint: Expand $(a+1)^2$. $\endgroup$ – Henning Makholm Sep 17 '16 at 18:12
  • 2
    $\begingroup$ This is a mere translation into formula of what has to be proved. What stucks you? $\endgroup$ – Bernard Sep 17 '16 at 18:12
  • $\begingroup$ the quantifiers is what throws me off. If the quantifiers weren't there then I would do a straight algebraic proof. $\endgroup$ – jdlucero7 Sep 17 '16 at 18:42
  • $\begingroup$ The quantifies say $a$ is given. You can find $b$ and $n$. You are kind of working backwords by starting with the $n^2$. If you start with $ab + 1 = n^2$ you need to work backwords and solve for $b$ in terms of $a$. $ab + 1 = n^2$ then $b = (n^2 - 1)/a = (n+1)(n-1)/a$. We need to argue that we can find $n$ where that works. Obviously we can let $n-1 = k*a$ or $n = k*a - 1$ and $b= k(ka-1)$. So for all $a$ then $a*(k(ka-1))+1=(ka-1)^2$ $\endgroup$ – fleablood Sep 17 '16 at 19:03
  • $\begingroup$ You do not need $n$ or $k$. You are looking for a good $b$ depending on $a$. If you try this, certainly $b=a+2$ comes to mind. Then $ab+1=a(a+2)+1=(a+1)^2$ is a square. $\endgroup$ – Dietrich Burde Sep 17 '16 at 19:17

$$a*b+1=n^2$$ $$a*b=n^2-1$$ $$a*b=(n-1)*(n+1)$$ From there, let $a$ or $b$ equal one of the two terms on the right side.


If you set $b=a+2$, then $ab+1=a(a+2)+1=a^2+2a+1=(a+1)^2$, as per Henning's comment.


The quantifiers say that $a$ is fixed and that you need to find some $b$ and $n$ in terms of $a$.

So if we want $ab + 1 = n^2$ we want $b = \frac{n^2 -1}a$ and $n = \pm\sqrt{ab+1}$.

The latter doesn't suggest any thing but $b = \frac{n^2 -1}a = \frac{(n-1)(n+1)}a$

So all we have to do is pick any $n$ so that $a|n\pm 1$. Why not simply $n = a-1$?

If $n = a - 1$ then $b = \frac{(n-1)(n+1)}a= \frac {(a-2)a}a = a-2$

So for every $a$ we can always find $b = a-2$ and $n = a-1$ so $ab + 1 = a(a-2) + 1 = a^2 - 2a +1 = (a-1)^2 = n^2$. Always. For all $a$.

These aren't the only ones. We could have found $n= ak\pm1$ and $b = k(ka \pm 2)$ so $ab + 1 = ak(ka \pm 2) = a^2k^2 \pm 2ak + 1 = (ak \pm 1)^2 = n^2$ for any $k$.


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