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Is the Following Proof Correct?

Theorem. Given that $\forall n\in\mathbf{N}(F_n = 2^{2^n}+1)$ show that the following is true $$\forall n\in{1,2,3...}\left(F_n = \prod_{j=0}^{n-1}F_j+2\right)$$ Proof. We construct the proof by recourse to Mathematical-Induction.

Basis-Step: For $n=1$ we have $F_1=F_0+2=(2^{2^0}+1)+2 = (2+1)+2=5$.

Inductive-Step: Assume now for an arbitrary $k\in\{1,2,3,...\}$ the following $$F_k = \left(\prod_{j=0}^{k-1}F_j+2\right)\tag{1}$$ consequently using $(1)$ we have $$\left(\prod_{j=0}^{k}F_j+2\right) = \left(\prod_{j=0}^{k-1}F_j\cdot F_k+2\right)=(F_k-2)\cdot F_k=(F_k)^{2}-2F_k+2\tag{2}$$ thus $(F_k)^{2}-2F_k+2 = (2^{2^k}+1)^2-2(2^k+1)+2$ expanding all terms we have $(2^{2^k}+1)^2-2(2^k+1)+2 = 2^{2^{k+1}}+2\cdot 2^{2^k}+1-2\cdot 2^{2^k}-2+2 = 2^{2^{k+1}}+1 = F_{k+1}$ thus $$F_{k+1}=\left(\prod_{j=0}^{k}F_j+2\right)\tag{3}$$

$\blacksquare$

Blue
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atifcppprogrammer
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2 Answers2

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Your proof is fine, but I think it's a little easier if you define the numbers $F_n$ by the recurrence $$F_n=F_0F_1\cdots F_{n-1}+2$$ and then prove the identity $F_n=2^{2^n}+1$ by induction.

Basis step: $$F_0=3=2^{2^0}+1$$ Inductive step: $$F_{n+1}=F_0\cdots F_{n-1}F_n+2$$ $$F_{n+1}-2=(F_0\cdots F_{n-1})F_n=(F_n-2)F_n=\left(2^{2^n}-1\right)\left(2^{2^n}+1\right)=\left(2^{2^n}\right)^2-1=2^{2^{n+1}}-1$$ $$F_{n+1}=2^{2^{n+1}}+1$$

bof
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  • But the inductive step needs to be proven for all numbers $k$, else (without proving) it can be a possible fallacious use of MI. If I am not wrong, you are assuming the inductive step without showing it to be true for all possible values of $k$. – jiten Dec 26 '17 at 02:02
  • Sorry, $k$ referred to the usual variable name in MI. You have used $n$. My intent was to to ask how is that assumption for inductive step, being true for all $n$ is taken up for the fermat numbers, $F_n$. In this case, do we need to 'implicitly' consider all possible values of $n$ for which the statement is true. If yes, then it satisfies me. Else, I feel it is a fallacious use of the MI technique. – jiten Dec 26 '17 at 03:10
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    I still don't understand your question. (1) I define $F_n$ recursively by athe equation $F_n=F_0F_1\cdots F_{n-1}+2.$ (2) I observe that $F_0=2^{2^0}+1.$ (3) In the inductive step, for arbitrary but fixed $n,$ I prove the implication $$F_n=2^{2^n}+1\implies F_{n+1}=2^{2^{n+1}}+1.$$ I suppose your question is about the inductive step. – bof Dec 26 '17 at 03:48
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    @jiten Assume $$F_n=2^{2^n}+1.\tag1$$ By definition of $F_{n+1}$ we have $$F_{n+1}-2=F_0F_1\cdots F_{n-1}F_n.\tag2$$ By definition of $F_n$ we have $$F_0F_1\cdots F_{n-1}=F_n-2.\tag3$$ Multiplying $(3)$ by $F_n$ we get $$F_0F_1\dots F_{n-1}F_n=(F_n-2)F_n.\tag4$$ By the inductive hypothesis $(1)$ we have $$(F_n-2)F_n=(2^{2^n}-1)(2^{2^n}+1).\tag5$$ By simple algebra $$(2^{2^n}-1)(2^{2^n}+1)=2^{2^{n+1}}-1.\tag6$$ From this chain of equalities we have $$F_{n+1}-2=2^{2^{n+1}}-1.\tag7$$ Adding $2$ to both sides of $(7)$ we get $$F_{n+1}=2^{2^{n+1}}+1.\tag8$$ Q.E.D. At which step you see a problem? – bof Dec 26 '17 at 04:09
  • I am unable to explain myself, was interested in details rather than dis-proving. Please wait for some time for comment that covers some needed details. – jiten Dec 26 '17 at 04:16
  • Please take me truly as a novice, and just explain me as a student that if you were to take the assumption of the 'inductive step' to be true - for this case -, then (i) do you rely on the substitution of n in the n+1 case, & that leading to failure/success as a measure of the formula being conjectured being false / true respectively; or (ii) you mentally, or on paper, or by literature work out all possible cases for the 'inductive step' to be true. I follow first approach. I feel it dis-satisfactory, but practical, as am unable to work out the truth of the inductive step beyond a few values. – jiten Dec 26 '17 at 04:40
  • Anyway, induction is for proving a conjecture. And the phrase "all possible values should hold for the inductive step" sounds weird , impractical, and directly obviating the need for the '(n+1th)' step. – jiten Dec 26 '17 at 04:44
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    I'm sorry but I find your comments very difficult to understand. And I don't know where your quotation "all possible values should hold for the inductive step" was taken from. Without context, I'm not sure what it even means. – bof Dec 26 '17 at 04:50
2

Looks good to me :)

NB - those are Fermat numbers, not Fermat primes.

Michael Hartley
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