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Let p(x) = x^2 +1 Then the failure function, when x =1, is 1 +2*k where k belongs to Z. Thus x =3,5,7,….when substituted in f(x) will yield failures i.e. composite numbers. Similarly when x=2, the failure function is 2 + 5*k.Thus 7,12, 17… , when substituted in f(x) yield failures i.e.composites congruent to 0 (mod(5)). ( Recall that our original definition of a failure is a compsite number.

Let f(x) be an exponential functionPlanetmathPlanetmath. Let our definition of a failure continue to be a composite number. Then x = x_0 + k*Eulerphi(f(x_0)) is a failure function. Here k belongs to N. In other words values of x generated by the above failure function, when substituted in f(x), will yield only failures ( composites ).

Let the exp. function be 2^x+7 =f(x). When n = 2, f(x)=11.Then x=2+10*k is a failure function. In other words values of x generated by this failure function, when substitued in f(x) will yield only failures (composites) exactly divisible by 11.

The whole object of the messages pertaining to this subject is to give a brief proof of the infinitude of primes with the form x^2 + 1, basing the proof on a) failure functions and b) the concept of proof by inevitablity of perpetual iteration. This will be furnished shortly.

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