Let _a_ + _b_ give the remainder _r_ (not unity); then _a_² ÷ _b_ gives the same remainder as _r__a_ + _b_, which (Prop. 4) cannot be _r_: let it be _s_. Then _a_ˢ ÷ _b_ gives the same remainder as _s__a_ ÷ _b_, which (Prop. 4) cannot be either _r_ or _s_, unless _s_ be 1: let it be _t_. Then _a_ᵗ ÷ _b_ gives the same remainder as _ta_ ÷ _b_; if _t_ be not 1, this cannot be either _r_, _s_, or _t_: let it be _u_. So we go on getting different remainders, until 1 occurs as a remainder; after which, at the next step, the remainder of _a_ ÷ _b_ is repeated. Now, 1 must come at last; for division by _b_ cannot give any remainders but 0, 1, 2, ... _b_- 1; and 0 never arrives (Prop. 3), so that as soon as _b_-2 _different_ remainders have occurred, no one of which is unity, the next, which must be different from all that precede, must be 1. If not before, then at _a_ᵇ⁻¹ we must have a remainder 1; after which the cycle will obviously be repeated. Thus, 7, 7², 7³, 7⁴, &c. will, when divided by 5, be found to give the remainders 2, 4, 3, 1, &c. PROP. 6. The difference of two _m_th powers is always divisible without remainder by the difference of the roots; or _a_ᵐ -_b_ᵐ is divisible by _a_-_b_; for _a_ᵐ - _b_ᵐ = _a_ᵐ - _a_ᵐ⁻¹_b_ + _a_ᵐ⁻¹_b_ - _b_ᵐ = _a_ᵐ⁻¹(_a_ - _b_) + _b_(_a_ᵐ⁻¹ - _b_ᵐ⁻¹) From which, if _aᵐ⁻¹_-_bᵐ⁻¹_ is divisible by _a_ -_b_, so is _a_ᵐ-_b_ᵐ. But _a_-_b_ is divisible by _a_-_b_; so therefore is _a_²- _b_²; so therefore is _a_³-_b_³; and so on. Therefore, if _a_ and _b_, divided by _c_, leave the same remainder, _a_² and _b_², _a_³ and _b_³, &c. severally divided by _c_, leave the same remainders; for this means that _a_-_b_ is divisible by _c_. But _a_ᵐ - _b_ᵐ is divisible by _a_-_b_, and therefore by every measure of _a_-_b_, or by _c_; but _a_ᵐ-_b_ᵐ cannot be divisible by _c_, unless _a_ᵐ and _b_ᵐ, severally divided by _c_, give the same remainder. PROP. 7. If _b_ be a prime number, and _a_ be not divisible by _b_, then _a_ᵇ and (_a_-1)ᵇ + 1 leave the same remainder when divided by _b_. This proposition cannot be proved here, as it requires a little more of algebra than the reader of this work possesses.[64] [64] Expand (_a_-1)ᵇ by the binomial theorem; shew that _when b is a prime number_ every coefficient which is not unity is divisible by _b_; and the proposition follows.