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Question Number 23050 by jazary last updated on 25/Oct/17

h o w c a n d e m o n s t r e d t h a t

\forall a, b, c \in N

a^{2} + b^{2} = c^{2} \Rightarrow

a b c \equiv 0 [60]

Answered by Rasheed.Sindhi last updated on 29/Oct/17

a b c \equiv 0 [60] \Rightarrow 60 ∣ a b c

\Rightarrow (4.3.5) ∣ a b c

\Rightarrow 4 ∣ a b c \land 3 ∣ a b c \land 5 ∣ a b c

^{∙} 4 ∣ a b c \Rightarrow 4 ∣ a b \lor 4 ∣ b c \lor 4 ∣ a c \dots \dots (i)

^{∙} 3 ∣ a b c \Rightarrow 3 ∣ a \lor 3 ∣ b \lor 3 ∣ c \dots \dots \dots . . (ii)

^{∙} 5 ∣ a b c \Rightarrow 5 ∣ a \lor 5 ∣ b \lor 5 ∣ c \dots \dots \dots . (iii)

We have to prove :

Case - 1 : (i) : 4 ∤ a b \land 4 ∤ b c \Rightarrow 4 ∣ a c

This can be proved like Case - 2

Case - 2 : (ii) : 3 ∤ a \land 3 ∤ b \Rightarrow 3 ∣ c

3 ∤ a \land 3 ∤ b means a and b or of

3 k + 1 or 3 k + 2 type

\Rightarrow a^{2} and b^{2} are of 3 k + 1 type

c = \sqrt{a^{2} + b^{2}} = \sqrt{(3 k + 1) + (3 l + 1)}

= \sqrt{3 m + 2}

But 3 m + 2 is not perfect square

∴ If c \in N \Rightarrow a^{2} + b^{2} is perfect square

\Rightarrow a^{2} and b^{2} are not both of type

3 k + 1 .

∴ a or b {aren}^{'} t of type 3 k + 1 or

3 k + 2 .

∴ a or b is 3 k - type

∴ 3 ∣ a or 3 ∣ b

∴ 3 ∣ a b c

Case - 3 : (iii) : 5 ∤ a \land 5 ∤ b \Rightarrow 5 ∣ c

This case can also be proved like Case - 2

4 ∣ a b c Case - 1

3 ∣ a b c Case - 2

5 ∣ a b c Case - 3

Hence 60 ∣ a b c

Or a b c \equiv 0 [60]