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Silk Road Mathematics Competition competition problems

Number theory Difficulty 5.1 AIME, harder Prove it Silk Road Mathematics Competition

Prove that for every prime number pp there exists infinitely many 4-tuples (x,y,z,t)(x, y, z, t) of pairwisely distinct positive integers such that the number
(x2+pt2)(y2+pt2)(z2+pt2) (x^2 + pt^2)(y^2 + pt^2)(z^2 + pt^2)
is a square of an integer.

Solution

Firstly, note that the equation x2−py2=1x^2 - py^2 = 1 has infinitely many solutions in positive integers. (Pell's equation)
Then for every prime number pp there exist infinitely many positive integers ss and tt such that s2−1=pt2s^2 - 1 = pt^2.
Putting x=s2−1x = s^2 - 1, y=s+1y = s + 1, z=s−1z = s - 1 we have
(x2+s2−1)(y2+s2−1)(z2+s2−1)==(s2−1)s2(s+1)(2s)(s−1)(2s)=((s2−1)2s2)2. (x^2 + s^2 - 1)(y^2 + s^2 - 1)(z^2 + s^2 - 1) = \\ = (s^2 - 1)s^2(s + 1)(2s)(s - 1)(2s) = ((s^2 - 1)2s^2)^2.
It is remained to check that x,y,z≠tx, y, z \ne t.
Note that x=s2−1=pt2≠tx = s^2 - 1 = pt^2 \ne t.
If y=s+1=ty = s + 1 = t then
(s−1)(s+1)=pt2=p(s+1)2⇒s−1=p(s+1)>s+1>s−1. (s-1)(s+1) = pt^2 = p(s+1)^2 \Rightarrow s-1 = p(s+1) > s+1 > s-1.
If z=s−1=tz = s - 1 = t, then
(s−1)(s+1)=pt2=p(s−1)2⇒s+1=p(s−1)≥2(s−1)>s+1, (s-1)(s+1) = pt^2 = p(s-1)^2 \Rightarrow s+1 = p(s-1) \ge 2(s-1) > s+1,
for s>3s > 3.

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