Let a,b,c be non zero integers,a≠c such that ac=a2+b2c2+b2Prove that a2+b2+c2 cannot be a prime number.
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Tags: number theory, prime
15.08.2024 10:56
https://math.stackexchange.com/questions/631198/why-cant-x2y2z2-be-prime?noredirect=1 https://artofproblemsolving.com/community/c4h1236079p6282199
15.08.2024 11:26
parmenides51 wrote: Let a,b,c be non zero integers,a≠c such that ac=a2+b2c2+d2Prove that a2+b2+c2 cannot be a prime number. (!) Instead of d is b. The official solution translated by ChatGPT: **Problem 2.** We have: ac=a2+b2b2+c2⟺(a−c)(b2−ac)=0.But a≠c and thus, b2=ac. In this case, we have: a2+b2+c2=a2+2ac+c2−b2=(a+c−b)(a+c+b).Since a2+b2+c2 is a natural number and a2+b2+c2>3, if a2+b2+c2 were a prime natural number, then we would have the following cases: 1. a+c−b=1 and a+c+b=a2+b2+c2, 2. a+c+b=1 and a+c−b=a2+b2+c2, 3. a+c−b=−1 and a+c+b=−a2−b2−c2, 4. a+c+b=−1 and a+c−b=−a2−b2−c2. In cases (1) and (2) we are led to: −2(a+c)+a2+b2+c2+1=0⟺(a−1)2+(c−1)2+b2=1and considering that b∈Z, we obtain b2=1 and a=c=1. In cases (3) and (4) we have: 2(a+c)+a2+b2+c2+1=0⟺(a+1)2+(c+1)2+b2=1and considering that b∈Z, we obtain b2=1 and a=c=−1. Therefore, in all four cases we contradict the hypothesis (a≠c), hence a2+b2+c2is a composite natural number.
15.08.2024 12:49
parmenides51 wrote: Let a,b,c be non zero integers,a≠c such that ac=a2+b2c2+d2Prove that a2+b2+c2 cannot be a prime number. Isn't it b2?
15.08.2024 12:57
@above, indeed wording had a typo