$Given$:
$A$, $B$, $C$, $E$, $F$ are distinct digits varying from $1$ to $9$.
$A$ is a Fibonacci number.
$BB$, $BC$, $EF$ are concatenated Numbers.
$Relationship$:
$(A*BB)*(BC)^2$ = $(EF)^2- B$
Deduce all the Digits.
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$\begingroup$ are BB, BC, EF fib no.s? $\endgroup$ – Omega Krypton Jun 26 '19 at 9:56
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$\begingroup$ Enough info is given to resolve $\endgroup$ – Uvc Jun 26 '19 at 9:57
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$\begingroup$ Do give the deductive reasoning which is very simple and straightforward. $\endgroup$ – Uvc Jun 26 '19 at 10:19
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Answer
$$(A,B,C,E,F) = (5,1,2,8,9)$$
Reasoning
$$A * BB \geq 22$$ $$ (EF)^2 - B \leq 98^2-1 = 9603$$ $$ \Rightarrow (BC)^2 \leq \frac{9603}{22} = \frac{873}{2} = 436 \frac{1}{2}$$ $$ \Rightarrow BC < 21 \Rightarrow B=1$$ $$ A*BB* (BC)^2 = A*11*(1C)^2 = (EF)^2-1 = (EF-1)(EF+1)$$ which means that either $EF-1$ or $EF+1$ is divisible by $11$ (since $11$ is prime).
This leaves the possibilities for $EF$ as being $23$, $32$, $34$, $43$, $45$, $54$, $56$, $65$, $67$, $76$, $78$, $87$, $89$ and $98$.
We can rule out those which are adjacent to numbers which are divisible by primes greater than $19$ since the left-hand side cannot have such a factor (that is $32$, $45$, $54$, $78$, $87$ and $98$) which leaves $8$ possibilities for $EF$ $\rightarrow$ $23$, $34$, $43$, $56$, $65$, $67$, $76$ and $89$.
Furthermore, $\frac{(EF-1)(EF+1)}{11}$ must be divisible by the square of a number $>11$.
Since $\gcd(EF-1, EF+1) \leq 2$, this means that either $EF-1$ or $EF+1$ is divisible by a square $>11^2$ (not possible) when $EF$ is even or that either $EF-1$ or $EF+1$ is divisible by an odd square when $EF$ is odd.
This is only true in one case, $EF=89$ and here, we find a solution.
$EF = 89 \Rightarrow (EF-1)(EF+1) = 2^4*3^2*5*11 \Rightarrow A*(1C)^2 = 2^4*3^2*5$ $$ \Rightarrow A=5, \,\,\,C=2$$
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$\begingroup$ Excellent deduction..famous Relationship can be seen from slight rearrangement of terms $\endgroup$ – Uvc Jun 26 '19 at 10:39
