#### Question

If m is a four-digit positive number, show there exists positive number n such that m-n is prime and mn is a perfect square number.

Let m-n = p, in which p is a prime number

m = n+p

Let k be integer such that mn = n(n+p) = k^2

n^2+np=k^2

4n^2+4np+p^2=4k^2+p^2

(2n+p)^2 - 4k^2 = p^2

Apply difference of squares formula

(2n+p+2k)(2n+p-2k) = p^2

Let a = 2n+p+2k, b = 2n+p-2k, then a\cdotp b = p^2

n = \dfrac{a+b-2p}{4}

Since p is a prime number, there'are only 2 pairs of factors for p^2, that is (p,p) and (1,p^2)

When a = b = p, n = 0, (cancel)

When a = 1, b=p^2

n = \dfrac{1}{4}(p-1)^2

m = n+p = p+ \dfrac{p^2-2p+1}{4} =\dfrac{1}{4}(p+1)^2

\because 1000 \leq m < 2010

1000 \leq \dfrac{1}{4}(p+1)^2 < 2010

64 \leq p +1 \leq 89

p could be prime numbers 67, 71, 73, 79, 83

Accordingly, m is 1156, 1296, 1369, 1600, 1764