I.S.I. Entrance 2018 Problem 7 | Bases, Exponents and Role reversals
Join Trial or Access Free ResourcesJoin Trial or Access Free ResourcesThis is a problem from ISI B.Stat-B.Math Entrance Exam 2018, Subjective Problem 7. It is based on Bases, Exponents and Role reversals.
Let $(a, b, c)$ are natural numbers such that $(a^{2}+b^{2}=c^{2})$ and $(c-b=1)$. Prove that
(i) a is odd.
(ii) b is divisible by 4
(iii) $( a^{b}+b^{a} )$ is divisible by c
Isolate a
Notice that \( a^2 = c^2 - b^2 = (c+b)(c-b) \)
But c - b = 1. Hence \( a^2 = c + b \). But c and b are consecutive numbers (after all their difference is 1!). Sum of two consecutive numbers is always odd (Why?).
Hence (a^2 ) is odd. This implies a is odd.
Eliminate c
Replace c by 1+b. We have \( a^2 + b^2 = ( 1+ b)^2 = 1 + b^2 + 2b \)
But that means \( a^2 = 1 + 2b \). This implies \( 2b = a^2 - 1 = (a-1)(a+1) \)
We already showed a is odd. Hence odd + 1, odd -1 are consecutive even numbers. Hence atleast one of them must be divisible by 4 implying their product must be divisible by 8 (as the other is divisible by at least 2).
Hence 2b equals something that is divisible by 8. This implies b is divisible by 4.
A bit of Modular Arithmetic
We already found that \( a^2 = c + b \).
We test this modulo c (Modular Arithmetic is a useful tool from Number Theory that you should definitely learn).
\( a^2 \equiv c + b \equiv b \mod c \) . But since b = c - 1, hence \( b \equiv -1 \mod c \). Hence \( a^2 \equiv -1 \mod c \)
Now recall that b is divisible by 8. Hence b/2 is even. This implies \( (a^2)^{\frac{b}{2}} \equiv (-1)^{\frac{b}{2}} = 1 \mod c \). Hence \(a^b \equiv 1 \mod c \)
On the other hand a is odd and b = c-1. Hence \( b^a \equiv (-1)^a \equiv -1 \mod c \)
Adding we get the final answer \( a^b + b^a \equiv 1 - 1 \equiv 0 \mod c \)
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