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Solution

We are solving n = dq + r and clearly the remainder must be less than the divisor; that is, r d.

If r q, then the terms will be ordered, d r q.

Let r = qx and d = qx², where x is the common ratio.

n = dq + r = q²x² + r = r² + r = r(r + 1)

But r(r + 1) can only be prime if r = 1, and as r q, q would be non-integer or zero.

Hence q r, and without loss of generality we shall suppose that d q r.

Let q = rx and d = rx².

n = dq + r = r²x³ + r = r(x³r + 1)

If r = 1 then n = x³ + 1. Therefore x³ must be integer, and as x is the common ratio of a geometric progression x 1.

But x³ + 1 = (x + 1)(x² x + 1) = (x + 1)(x(x 1) + 1), and as x 2, it follows that x(x 1) + 1 3. Hence p would be the product of two numbers greater than two and could not be prime.

So let us consider the case where r 1.

If x is integer then p = r(x³r + 1) will be the product of two numbers greater than one and cannot be prime.

Next, consider x being non-integer; let x = a/b where GCD(a, b) = 1.

n = r(x³r + 1) = r(a³r/b³ + 1) = (r/b)(a³r/b² + b)

But d = rx² = ra²/b² and as d is integer it follows that b² divides r and we can see that a³r/b² + b will be integer greater than one in value.

Let r/b² = c r/b = bc, and as b 2, r/b 2.

Hence n would again be the product of two integers greater than one and cannot be prime.    Q. E. D.