Use the following Lemma: let p be a prime number, and suppose p divides the numb

Question

Use the following Lemma: let p be a prime number, and suppose p divides the number ab. Then either p divides a nor p divides b (or p divides both a and b).

Let s, t be odd integers with s > t 1 and gcd (s, t) = 1. Prove that the three numbers a = st, b = s2 - t2/2 c = s2 + t2/2. are pairwise relatively prime: that is, each pair of them is relatively prime. We needed this fact to finish classifying primitive Pythagorean triples, (Hint, similarly to 5., suppose that there is a common prime factor

Explanation / Answer

(b,c) let b,c have a common prime factor 'p' (s^2-t^2)/2=k1*p (s^2+t^2)/2=k2*p adding both (2S^2)/2=(k1+k2)p s*s=(k1+k2)p using lemma p should s ie p|s now subtrac both (2t^2)/2=(k2-k1)p t*t=(k2-k1)p using lemma p should t ie p|t contradicting s,t are reletively primes so,b,c are reletively primes (a,b) let a,b have a common prime factor 'p' (s^2-t^2)/2=k1*p st=k2*p (using lemma p|s or p|t p divides both s,t) if p divides both s,t we are done (i)suppose p|s then s=k3*p (s^2-t^2)/2=k1*p ((k2*p)^2-t^2)/2=k1*p t^2=k2*k2*p*p - 2k1*p t^2=(k2*k2*p - 2k1)*p (t^2)/p=(k2*k2*p - 2k1) so using lemma p|t ,so,contradiction (ii)suppose p|t then t=k3*p (-(k2*p)^2+s^2)/2=k1*p s^2=k2*k2*p*p + 2k1*p s^2=(k2*k2*p + 2k1)*p so using lemma p|s contradicting p divides both s,t (a,c) let a,c have a common prime factor 'p' (s^2+t^2)/2=k1*p st=k2*p (using lemma p|s or p|t p divides both s,t) if p divides both s,t we are done (i)suppose p|s then s=k3*p (s^2+t^2)/2=k1*p ((k2*p)^2+t^2)/2=k1*p t^2= -k2*k2*p*p + 2k1*p t^2= (-k2*k2*p + 2k1)*p so using lemma p|t contradiction (ii)suppose p|t then t=k3*p ((k2*p)^2+s^2)/2=k1*p s^2= -k2*k2*p*p + 2k1*p s^2= (-k2*k2*p + 2k1)*p so using lemma p|s contradicting p divides both s,t all are paiwise relatively prime

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