# Understand the problem

Show that for each non-negative integer $n$ there are unique non-negative integers $x$ and $y$ such that we have
$$n=\frac{(x+y)^2+3x+y}{2}.$$
[/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="4.0" hover_enabled="0"]Germany MO, 2019, Problem 4 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]4/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="on"]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"]

[/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" hover_enabled="0"][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="4.0" hover_enabled="0"]If you observe the expression $\frac{(x+y)^2 + 3x + y}{2}$. It has two free variables. Now, if let's play with the expression to make it a little more symmetric and easy to work with and make some meaning out of it. Try to remove the asymmetry of x and write it in a much more symmetric form in x and y.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]

Observe that  $\frac{(x+y)^2 + 3x + y}{2} = \frac{(x+y)^2 + (x+y)}{2} + x$ is the expression what we get after bringing in the symmetry. Now, factorize it and see what we are looking for is $n = \frac{(x+y)(x+y+1)}{2} + x$.  Can you guess anything about the expression $\frac{(x+y)(x+y+1)}{2}$.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]

$\frac{(k)(k+1)}{2}$  is the sum of the first k natural numbers.  So, now the idea is that somehow you are taking the first k natural numbers and adding another number x to it to make any number. Can you get the final logic?   [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Now, observe that sum of the first k numbers is increasing with k.  Now, take any number say 17. Now observe that 17 lies between 15 and 21.  This means that 17 = 15 + 2 = ( 1 +2 + 3 +4 + 5) +  2. So, this is the idea that k = x+y is the largest number such that n is greater than or equal to the ( 1 + 2 + ... + k ) and observe that we may need something like 2 in case of n = 17. Call that x and obviously $k \geq x$. Hence, define y = k - x. So, for n = 33 = ( 1 + 2 + ... +  7 ) + 5.  Hence k = x + y = 7, x = 5, y = 2.  This is the whole idea. QED.  [/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"]

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