that satisfy the equation :
[/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"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Greece MO 2019, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Challenges and Thrills of Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]
and 
where 
.We have
And since 
we must have 
and 
.Now, we have to find the solutions from these equations.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Observe both the equations are of the form \(x^a = y^b\).The idea is that due to the prime factorization theorem, we can specify that \(x^a = y^b\) leads to a special form of the x and y.Claim: If 
for some 
naturals , then there exists a natural 
such that 
and 
where 
and 
.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]Proof of Claim:Consider the prime factorization theorem of the x and y in \(x^a = y^b\). Observe that it implies x and y must have same set of primes by the prime factorization theorem.Let x and y contain the primes \(p_1, p_2, ..., p_k\). Let \( x = \prod_{i=1}^{k} x_i\) and \( y = \prod_{i=1}^{k} y_i\).The above equation implies that \( a.x_i = b.y_i \). This implies that \( y_i = n.c \) and \( x_i = m.c \), where c is a natural number. Hence \( x = z^m, y = z^n\).[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Using the intuitive claim,
, there exists a such that 
, if 
we have 
and continuing like this , 
is unbounded ,contradiction.So , we must have 
wich means 
and from we have 
So, the solutions come out to be 
and 
, where 
is any positive integer.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]
