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I am trying to find a way to describe all integer values of $x$ for which the following holds true:

$\sqrt[2]{(1/2) * x * (x - 1) + (1/4)} + (1/2)\in \mathbb{Z}$

Noting that this can be equivalently stated as:

$2*\sqrt[2]{(1/2) * x * (x - 1) + (1/4)} \equiv 1 (mod 2)$

I am aware of the fact that, via Hensel's Lemma, if I can make the left-hand side a polynomial with integer coefficients, then the solutions can easily be found. Thus, I'm trying to manipulate this congruence to make it a polynomial congruent to $0$ modulo a prime. I believed that perhaps the following was equivalent to the above expression:

$4*((1/2)*x*(x-1)+(1/4)) \equiv 1 (mod 4)$

(I.e., just squaring both sides of the congruence as well as the modulo.)

Which could then trivially be manipulated to show that:

$2*x*(x-1) \equiv 0 (mod 4)$

However, it is easy to see that this is no longer equivalent by a simple counter-example (e.g., the first statement fails for $x = 121$, whereas the second statement is true). Squaring both sides of a congruence (at least in this way) doesn't seem to make any sense.

Is there a technique I can use to remove the radical from this congruence, or am I going about this the wrong way? I would prefer hints or general techniques, rather than an explicit solution to this particular problem.

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It seems the following.

Usually congruences are used for integer numbers, not for the reals. So solving your problem we should find an integer $y$ such that $x(x-1)/2+1/4=(y-1/2)^2$, that is $2(2y-1)^2-(2x-1)^2=1$. This is a partial case of minus Pell’s equation.

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  • $\begingroup$ This is great. I'm not totally convinced working with modulo arithmetic couldn't work, but reducing this problem to Pell's equation is honestly a more elegant solution anyway. Thanks! $\endgroup$ – CmdrMoozy Dec 11 '13 at 16:36

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