Electronic – Why does the crystal resonate at 4 times the specified frequency

If the gain of a system at every particular delay is constant, the system will produce oscillations with those periods which have a gain of one. At periods where the gain exceeds one, the strength of the oscillations will grow unless or until something causes the gain falls to below one. If there were one frequency where the gain stabilized at exactly one, and it were less than one at all other frequencies, the circuit would produce a sine wave at the frequency in question. The wave would be a sine wave because any other type of wave would have content at frequencies where the gain is less than one.

Note that in practice, many types of oscillating circuits have gains which vary widely during the course of each "cycle". Such variations make it very difficult to predict analytically the frequency content of their output. Because there is a very fine line between having oscillations die down to nothing, and having oscillations grow without bound, even circuits which are intended to produce sine waves generally end up having a gain which is sometimes greater than 1 and sometimes less than one, though ideally there's a gain control mechanism that will try to settle on the right value.

Incidentally, some circuits use incandescent light bulbs for that purpose, since their resistance varies with temperature. If the power fed to a light bulb is proportional to the strength of an oscillator's signal, and if an increase in resistance will cause a reduction in gain, then the light bulb's temperature will tend to reach an equilibrium where the gain is 1. If the frequency in question is fast enough, the light bulb will only heat up or cool down a little bit during each cycle, allowing reasonably-clean sine waves to be generated.

Addendum

Rather than using the term "constant gain", it may be more helpful to use the term "linear circuit". To borrow an analogy from a magazine I read some years back, comparing "linear circuits" to "non-linear circuits" is like comparing "kangaroo biology" to "non-kangaroo biology"; linear circuits are a particular subcategory of circuits, and non-linear circuits are everything else.

A one-input one-output linear black box is one which takes an input signal and produces an output signal, with the characteristic that if F(x) represents the signal produced by the box when it is fed signal x, and if A and B are two input signals, then F(A+B) will equal F(A)+F(B). There are many kinds of things a linear black box can do to a signal, but all must obey the above criterion. The output produced by a linear black box when given a combination of many different frequency signals will be the sum of the outputs that would be produced for each frequency in isolation.

The behavior of many practical circuits is close to that of a one-input one-output linear black box. Since any wave other than a sine wave is a combination of sine waves at multiple frequencies, for a circuit to oscillate with anything other than a sine wave, there must be multiple frequencies which, if fed in individually, would cause the output to precisely match the input in phase and amplitude. While it is certainly possible to construct such circuits, most practical circuits will only have one frequency where such behavior will occur.

As I see, no answer was accepted. Let me offer another answer.

Most modern ICs use so-called Pierce Oscillator to generate stable clocks using crystals. Here is the main circuit configuration:

As one can see, the circuit is not symmetrical: the right side is output of some driver (usually designated as XO), and the left side is input to an inverting amplifier (usually designated as XI). Therefore it is relatively safe to probe the XO (output) end, provided that the probe has a relatively high impedance. An usual 1:10 passive probe with 1M input impedance should do the job. In practice, the output driver in circuit amplifier is made intentionally weak, typically no more than 1mA load capability, to prevent Xtal from being overddriven, but 1mA should be well enough to drive a 1M scope probe.

The probe tip capacitance may shift the frequency of oscillation by 20-50ppm, since it will change the circuit tune-up (Xtal load, C1 in series with C2). However, the probe load on XO should not break oscillations, unless the entire circuit is too marginal and does not meet the stability criteria (negative impedance of amplifier should be 3-5 times more than Xtal ESR). If the probe does this, consider the Xtal test as fail.

One should never try to probe the XI input, maybe only with a 100 MOhm probe, and only for curioucity. The reason is not in the tip capacitance (2-8-12pF or wahtever), but in inflicting a DC shift on XI pin due to finite probe impedance. The Pierce oscillator is a very delicate non-linear circuit, and it has a very important DC feedback component R1, which effectively adjusts the input DC level to the point of maximum amplification, usually about half way from ground to Vcc. The component R1 is usually 1MOhm and above, and the oscillations become centered at self-selected DC point. Attaching even a 10MOhm probe shifs this point down, amplification drops, and oscillations die.

And, of course, the best way to test for oscillations is not to touch it with probes, but to have an internal buffer with output to some otherGPIO test pin.