Test of Mathematics Solution Subjective 43-Integer Root

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 43 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Show that the equation $ x^3 + 7x - 14(n^2 +1) = 0 $ has no integral root for any integer n.

Solution:

We note that $ 14(n^2 +1) - 7x = x^3 $ implies $ x^3 $ is divisible by 7. This implies x is divisible by 7 (as 7 is a prime number). Suppose x= 7x'. Hence we can rewrite the given equation as:

$ 7^3 x'^3 + 7 \times 7 x' - 14 (n^2 +1 ) = 0 $.

Cancelling out a 7 we have $ 7^2 {x'}^3 + 7{x'} = 2(n^2 +1) $. Since 7 divides left hand side, it must also divide the right hand side. Since 7 cannot divide 2, it must divide $ n^2 + 1 $ as 7 and 2 are coprime.  Note that 7 cannot divide $ n^2 +1 $ as square of a number always gives remainder 0, 1, 4, 2 when divided by 7 and never 6. But if $ n^2 + 1 $ is divisible by 7 then $ n^2 $ must give remainder 6 when divided by 7.  Hence contradiction.

Necessary Lemma: square of a number always gives remainder 0, 1, 4, 2 when divided by 7

$ n \equiv 0, \pm 1 , \pm 2 , \pm 3 \mod 7\Rightarrow n^2 \equiv 0, 1, 4, 9 (=2) \mod 7 $

Key Ideas: Modular Arithmetic

How are Bezout's Theorem and Inverse related? - Number Theory

The inverse of a number (modulo some specific integer) is inherently related to GCD (Greatest Common Divisor). Euclidean Algorithm and Bezout's Theorem forms the bridge between these ideas. We explore them in a very lucid manner.

Belarus MO 2018 Problem 10.5 - Number Theory

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Find all positive integers $n$ such that equation $$3a^2-b^2=2018^n$$has a solution in integers $a$ and $b$.

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Start with hints

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Belarus MO 2018 Problem 10.5 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.26.6" open="on"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.26.6" open="off"]5/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.26.6" open="off"]An Introduction to Number Theory [/et_pb_accordion_item][/et_pb_accordion][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.26.6" 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="3.26.6"]Let's check for n = 1. Observe that a = 27, b = 13 gives a solutions for n = 1. What about higher degrees? Can we use this information?  [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.26.6"]Does it work for n = 2? Let's prove something general! Prove that for a, b to have solutions, n must be odd.  [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.26.6"]If n is even, Take $\pmod{3}$ to see that $-b^2\equiv 1\pmod{3}$, which has no integer solutions in $b$.  Hence, n must be odd. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.26.6"]Well now take n odd. Say $n=2m+1$ for some positive integer $m$. Then, the solution $(a,b)=(27\times 2018^m, 13\times 2018^m)$ exists and works. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.26.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Watch video

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Connected Program at Cheenta

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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%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Similar Problems

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RMO 2018 Tamil Nadu Problem 3 - Nonlinear Diophantine Equation

RMO 2018 Tamil Nadu Problem 3 is from Number Theory. We present sequential hints for this problem. Do not read all hints at one go. Try it yourself.

Problem

Show that there are infinitely many 4-tuples (a, b, c, d) of natural numbers such that $a^3 + b^4 + c^5 = d^7$.

Key ideas you will need to solve this problem

Also see

Advanced Math Olympiad Program

Hint 1: Powers of 3

Powers of 3 have a very interesting property: $$ 3^n + 3^n + 3^n = 3^{n+1} $$

This simple observation is the key to this problem.

Hint 2: Expressing \( 3^n \) in multiple ways.

We want $ a^3 = 3^n, b^4 = 3^n, c^5 = 3^n $. Clearly n needs to be a multiple of 3, 4 and 5. For example, $3 \times 4 \times 5 = 60 $ may work. That is $$ 3^{60} + 3^{60} + 3^{60} = (3^{20})^3 + (3^{15})^4 + (3^{12})^5 $$

Hence in this case $$ a = 3^{20}, b = 3^{15}, c = 3^{12} $$

This will work for any multiple of 60. Suppose 60k is a multiple of 60 (that is k is any integer). Then we will have $$ a = 3^{20k}, b = 3^{15k}, c = 3^{12k} $$

Hint 3: We need 60k +1 to be a multiple of 7

Notice that $$ 3^{60k} + 3^{60k} + 3^{60k} = (3^{20k})^3 + (3^{15k})^4 + (3^{12k})^5 = 3^{60k+1} $$

We need $$ 3^{60k+1}  = d^7 $$

That is 60k +1 needs to be a multiple of 7. In terms of modular arithmetic we want $$ 60k + 1 \equiv 0 \mod 7 $$

$60 \equiv 4 \mod 7 \  \Rightarrow 60k + 1 \equiv 4k +1 \mod 7 \ \Rightarrow 4k \equiv - 1 \equiv 6 \mod 7$

This is where we will use the notion of inverse of a number modulo 7. Inverse of 4 modulo 7 is 2. This is because $ 4 \times 2 = 8 \equiv 1 \mod 7 $. The Bezout's theorem guarantees existence of inverse of 4 modulo 7. (Look into the reference at the end of this discussion if you do not know these ideas).

$4k \equiv 6 \mod 7 \ \Rightarrow 2\times 4k \equiv 2\times 6 \mod 7 \ \Rightarrow k \equiv 5 mod 7 $

Hence k = 7k' + 5 is suitable for our purpose.

Since there are infinitely many such integers with have infinitely many 4 tuples that will work.

Illustration: For k' = 0, k = 5. Therefore $60 \times 5 = 300$ should work. And it does: $$ 3^{300} + 3^{300} + 3^{300} = (3^{100})^3 + (3^{75})^4 + (3^{60})^5 = 3^{301} = (3^{43})^7 $$

Reference:

Also see

RMO 2018 Tamil Nadu Region

Test of Mathematics Solution Subjective 35 - Divisibility by 16

Test of Mathematics at the 10+2 Level

Test of Mathematics Solution Subjective 35 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

(a) Prove that, for any odd integer n, $ n^4 $ when divided by $16$ always leaves remainder $1$.

(b) Hence or otherwise show that we cannot find integers $n_1 , n_2 , ... , n_8 $ such that $n_1^4 + n_2^4 + ... + n_8^4 = 1993 $.

Solution

For part (a) we consider $n = 2k +1$ and expand it's fourth power binomially to get $ (2k+1)^4 = {{4} \choose {0}} (2k)^4 +{{4} \choose {1}} (2k)^3 + {{4} \choose {2}} (2k)^2 + {{4} \choose {3}} (2k)^1 + {{4} \choose {4}} (2k)^0 = 16k^4 + 32k^3 + 24k^2 + 8k + 1 $

Now $24k^2 + 8k = 8k(3k+1) $ ; if $k$ is even then $8k$ is divisible by $16$ and if $k$ is odd $3k+1$ is even and product of $8k$ and $3k+1$ is divisible by $16$. Since $16k^4 + 32 k^3 $ is already divisible by $16$ we conclude $ (2k+1)^4 $ when divided by $16$ gives $1$ as remainder.

For part (b) we note that $1993$ when divided by $16$, produces $9$ as the remainder. Each of the eight of fourth powers when divided by $16$ produces either $0$ (when $n_i $ is even) or $1$ (when $n_i $ is odd using part (a)) as remainder. Thus they can add up to at most $8$ (modulo $16$) hence can never be equal to $9$ (which $1993$ is modulo $16$).