Question

Let p >2 be a prime. Prove that there exists at least one positive integer n such that n^2+np is a perfect square.

Collected in the board: Perfect square integer

Steven Zheng posted 1 month ago


Answer 1

Let m be an integer such that

n^2+np=m^2

4n^2+4np+p^2-p^2=4m^2

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

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

Let a=2n+p-2m, b=2n+p+2m

Addition of the two equations gives

n=\dfrac{a+b-2p}{4}
(1)

And

a\cdotp b=p^2
(2)

Since p is a prime number, there are only 2 pairs of divisors for p^2, (p,p), (1,p^2)

When a=b=p, n = \dfrac{p+p-2p}{4}=0

When a = 1, b=p^2,

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

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

Verify:

When p=3, n = 1

n^2+np = 1+3 = 4, m = 2

When p = 5, n = 4

n^2+np = 16+20 = 36, m = 6

When p = 7, n = 9

n^2+np = 81+9\times 7 = 144 , m = 12

Steven Zheng posted 1 month ago


Answer 2

n^2+np =n(n+p)

If n^2+np is perfect square, n > 2, both n and n+p are perfect squares.

Let n = a^2, n+p = b^2

p = b^2-n=b^2-a^2 = (b-a)(b+a)

Since p is a prime number, it has only two factors, 1 and itself.

Let b-a = 1, then b+a = p

a = \dfrac{p-1}{2}

Therefore

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

Steven Zheng posted 1 month ago

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