If a,b,c are different from 0, a is different from c and (a2+b2)/(b2+c2)=a/c. Prove a2+b2+c2 cannot be a composite number

Hi Yugi,I'm a little unclear on your formatting - is your question:Suppose a,b,c are some non-zero integers, with a ~= c and (a^2+b^2)/(b^2+c^2) = (a/c). Show that (a^2+b^2+c^2) cannot be composite (i.e. is always prime). If so, we cannot prove this as this is false. Consider the case a=2, b=4, c=8. Then (a^2+b^2)/(b^2+c^2)= (4+16)/(16+64)=(20/80)=(2/8)=(a/c) and the sum of squares a^2+b^2+c^2 is 4+16+64=84, which is composite.
Tom H.
28 July 2017
it can be shown that there is an infinite set of numbers satisfying the first equation but for which a2+b2+c2 is composite. For example it is true for a=2 b=4 and c = 8 values.  It looks like some conditions are missing in statement
14 August 2017
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