# Resolve this Fibonacci Relationship

$$Given$$:

$$A$$, $$B$$, $$C$$, $$E$$, $$F$$ 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.

• are BB, BC, EF fib no.s? – Omega Krypton Jun 26 '19 at 9:56
• Enough info is given to resolve – Uvc Jun 26 '19 at 9:57
• Do give the deductive reasoning which is very simple and straightforward. – Uvc Jun 26 '19 at 10:19

$$(A,B,C,E,F) = (5,1,2,8,9)$$

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.

$$EF = 89 \Rightarrow (EF-1)(EF+1) = 2^4*3^2*5*11 \Rightarrow A*(1C)^2 = 2^4*3^2*5$$ $$\Rightarrow A=5, \,\,\,C=2$$

• Excellent deduction..famous Relationship can be seen from slight rearrangement of terms – Uvc Jun 26 '19 at 10:39