You could use a simple series resistor, but to compute the right resistor you'll need to know how much current your load draws. R = E/I. In this case, the E
we care about is the voltage drop we want (1 volt). Plug in the current the load will draw, and compute the resistance you need. There are serious problems here though. First of all, the voltage seen by your load will vary depending on how much current it draws, so unless its current draw is nearly constant, this approach simply can't work at all. Second, all the current flowing into the load also flows through the resistor, so unless its current draw is extremely low, your resistor is going to be dissipating a lot of energy.
To make it work regardless of the current drawn by the load, you could (at least theoretically) use a voltage divider:
simulate this circuit – Schematic created using CircuitLab
Even with a voltage divider, however, you want to design it for approximately the current your load will draw. The ratio of the resistors determines the output voltage, but all the current going to the load has to flow through the upper resistor, so with the values I used here, you have 5 volts through 100 ohms, giving a maximum of 5/100 = 50 ma.
Reducing the resistor values will increase the maximum current you can draw. So if we used 1 ohm and 4 ohms, the load could draw up to 5 amps.
But if we do that, it means that even when the load isn't drawing anything, the source is seeing a load of only 5 ohms, so it'll try to draw 1 amp. Again, our resistors are dissipating a lot of energy, which means you'll need big, expensive power resistors, and you're simply wasting a ridiculous amount of energy.
So, a voltage divider can maintain a (reasonably) constant voltage across changes in current draw, but you still want to choose the largest resistance values you can that will still supply the current you need. This will minimize the current you're constantly drawing from the source.
A voltage divider can work reasonably well as long as you only need really low current or if you only need to reduce the voltage by a tiny margin. But if you need to reduce voltage much and/or draw much current, you'll end up with a lot of energy being burned in the resistors themselves. It's particularly bad if your load's current draw varies widely, so you need to use low-value resistors to support the necessary current, so you're wasting a lot of energy all the time just to support a large current draw once in a while.
Unless your current requirements are extremely low, you usually want a buck converter, not resistors.