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August 29, 2018

Bases, Exponents and Role reversals (I.S.I. Entrance 2018)

[et_pb_section fb_built="1" admin_label="Blog Hero" _builder_version="3.22" use_background_color_gradient="on" background_color_gradient_start="rgba(114,114,255,0.24)" background_color_gradient_end="#ffffff" background_blend="multiply" custom_padding="0|0px|0|0px|false|false" animation_style="slide" animation_direction="top" animation_intensity_slide="2%" locked="off"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" custom_margin="|||" custom_padding="27px|0px|27px|0px" custom_width_px="1280px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_text_color="#474ab6" text_line_height="1.9em" background_size="initial" background_position="top_left" background_repeat="repeat" text_orientation="center" max_width="540px" module_alignment="center" locked="off"]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
[/et_pb_text][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" admin_label="Blog" _builder_version="3.22" custom_margin="|||" custom_padding="0px|0px|21px|0px|false|false"][et_pb_row _builder_version="3.25" background_size="initial" background_position="top_left" background_repeat="repeat" max_width="960px" custom_padding="0|0px|24px|0px|false|false" use_custom_width="on" custom_width_px="960px"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_tabs _builder_version="3.12.2"][et_pb_tab title="Hint 1 - Isolate a" _builder_version="3.12.2"]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.
[/et_pb_tab][et_pb_tab title="Hint 2 - Eliminate c" _builder_version="3.12.2"]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.
[/et_pb_tab][et_pb_tab title="Hint 3 - A bit of Modular Arithmetic" _builder_version="3.12.2"]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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