How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Pythagorean Triple, ISI Subjective Entrance 2018, Problem 7

## Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] Let $a,b,c \in \mathbb{N}$ be 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_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.3.4" hover_enabled="0"]
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 7.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]8 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="off"]

'Elementary Number Theory' by David M. Burton

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

$a^2+b^2=c^2=(b+1)^2=b^2+2b+1$ $\Rightarrow a^2=2b+1$ Which implies $a^2$ is odd integer. $\Rightarrow a$ is also an odd integer =$2k+1$ (say).

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

$(2k+1)^2+b^2=c^2$ $\Rightarrow 4k^2+4k+1+b^2=b^2+2b+1$ $\Rightarrow b=2k(k+1)$. Now $k(k+1)$ is always even =$2l$ (say). Therefore, $b=4l$, i.e. $b$ is divisible by 4.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

$a^b+b^a=a^{4l}+b^{2k+1}=(a^2)^{2l}+(c-1)^{2k+1}$ =$(2b+1)^{2l}+(c-1)^{2k+1}$ =$(2c-1)^{2l}+(c-1)^{2k+1}.$

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Now $2l$ is even , therefore , $(2c-1)^{2l}$ is of the form : $2cp+1$ where $p \in \mathbb{N}$. And $2k+1$ is odd , therefore $(c-1)^{2k+1}$ is of the form : $cq-1$, where $q \in \mathbb{N}$. Therefore $a^b+b^a=(2cp+1)+(cq-1)=c\cdot (2p+q)$. $\Rightarrow c|a^b+b^a$.

## Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]