Three satellites fix your position in space, but your phone's clock is not perfect, so a fourth is needed to solve for the clock error too.
You may have learnt that three distances are enough to fix a point in three dimensions. That is true if you know the distances exactly. A navigation receiver does not.
The receiver measures distance by timing a signal. To do that it compares the time stamp from the satellite with its own clock. A phone's clock is far less precise than the clocks on navigation satellites. If it is off by even a tiny fraction of a second, every distance comes out wrong.
So the receiver treats its own clock error as one more unknown. Now there are four: east–west position, north–south position, height, and the clock offset. ESA's Navipedia puts it plainly: there are four unknown parameters, so measurements from at least four satellites are needed.
The nice part is that the clock error is the same for every satellite at that moment. The receiver finds the one clock correction that makes all four distances meet at a single point.
With five, eight or twenty satellites in view, the receiver has more equations than unknowns. It then picks the best-fitting answer using least squares, which smooths out errors from any one signal.
This is why systems are designed for plenty of satellites. GPS uses 24 baseline slots so that at least four are visible from almost anywhere. India's NavIC uses seven satellites placed high over the Indian region: three geostationary and four in inclined geosynchronous orbits. Modern phones listen to several systems together, so they usually see far more than four.
Count how many NavIC satellites are above India at the same time: a fix needs at least four in view. Count the NavIC satellites above India
Write the problem for satellite j at known position (x_j, y_j, z_j):
R_j = √((x_j − x)² + (y_j − y)² + (z_j − z)²) + c × b
where R_j is the measured pseudorange and b is the receiver clock bias in seconds. The unknowns are x, y, z and b: four unknowns, so four independent equations are the minimum.
How big is the clock problem? Suppose the receiver's clock is off by 1 ms. That error (a thousandth of a second) becomes:
c × b = 299,792 km/s × 0.001 s ≈ 300 km
Every one of the four pseudoranges is wrong by the same 300 km. If the receiver ignored b and used only three satellites, the spheres would meet at the wrong point. Treating b as an unknown fixes that.
Worked idea in one dimension: imagine two beacons on a straight road, 10 km apart, and you are somewhere between them. With a perfect clock, one beacon is enough. With an unknown clock bias, distance to beacon A is d_A + e and to beacon B is d_B + e, where e is the same error. Since d_A + d_B = 10 km, adding the two measurements gives 10 + 2e, so you can find e, then your position. Each extra dimension adds one more unknown, so 1D needs 2 beacons, 2D needs 3, and 3D needs 4.
With more than four satellites, the receiver solves the over-determined set by least squares, as Navipedia describes.
The receiver's clock offset from system time.
About 300 m, because 3 × 10⁸ m/s × 10⁻⁶ s = 300 m.
Extra satellites give more equations than unknowns, and a least-squares solution reduces the effect of errors in any one signal.