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Landscape evolution models often utilize the stream power incision model to simulate river incision: <i>E</i> = <i>KA<sup>m</sup>S<sup>n</sup></i>, where <i>E</i> = vertical incision rate, <i>K</i> = erodibility constant, <i>A</i> =  upstream drainage area, <i>S</i> = channel gradient, and m and n are exponents. This simple but useful law has been employed with an imposed rock uplift rate to gain insight into steady-state landscapes. The most common choice of exponents satisfies <i>m/n</i> = 0.5; indeed, this ratio has been deemed to yield the “optimal channel network.” Yet all models have limitations. Here, we show that when hillslope diffusion (which operates only at small scales) is neglected, the choice <i>m/n</i> = 0.5 yields a curiously unrealistic result: the predicted landscape is invariant to horizontal stretching. That is, the steady-state landscape for a 1 m<sup>2</sup> horizontal domain can be stretched so that it is identical to the corresponding landscape for a 100 km<sup>2</sup> domain.