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Difference between revisions of "1984 IMO Problems/Problem 2"

(Created page with "==Problem== Find one pair of positive integers <math>a,b</math> such that <math>ab(a+b)</math> is not divisible by <math>7</math>, but <math>(a+b)^7-a^7-b^7</math> is divisibl...")
 
(→‎Solution 2)
 
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Find one pair of positive integers <math>a,b</math> such that <math>ab(a+b)</math> is not divisible by <math>7</math>, but <math>(a+b)^7-a^7-b^7</math> is divisible by <math>7^7</math>.
 
Find one pair of positive integers <math>a,b</math> such that <math>ab(a+b)</math> is not divisible by <math>7</math>, but <math>(a+b)^7-a^7-b^7</math> is divisible by <math>7^7</math>.
  
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==Solution==
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== Solution 1 ==
 
So we want <math>7 \nmid ab(a+b)</math> and <math>7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2</math>, so we want <math>7^3 | a^2+ab+b^2</math>.
 
So we want <math>7 \nmid ab(a+b)</math> and <math>7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2</math>, so we want <math>7^3 | a^2+ab+b^2</math>.
 
Now take e.g. <math>a=2,b=1</math> and get <math>7|a^2+ab+b^2</math>. Now by some standard methods like Hensels Lemma (used to the polynomial <math>x^2+x+1</math>, so <math>b</math> seen as constant from now) we get also some <math>\overline{a}</math> with <math>7^3 | \overline{a}^2+\overline{a}b+b^2</math> and <math>\overline{a} \equiv a \equiv 2 \mod 7</math>, so <math>7\nmid \overline{a}b(\overline{a}+b)</math> and we are done. (in this case it gives <math>\overline{a}=325</math>)
 
Now take e.g. <math>a=2,b=1</math> and get <math>7|a^2+ab+b^2</math>. Now by some standard methods like Hensels Lemma (used to the polynomial <math>x^2+x+1</math>, so <math>b</math> seen as constant from now) we get also some <math>\overline{a}</math> with <math>7^3 | \overline{a}^2+\overline{a}b+b^2</math> and <math>\overline{a} \equiv a \equiv 2 \mod 7</math>, so <math>7\nmid \overline{a}b(\overline{a}+b)</math> and we are done. (in this case it gives <math>\overline{a}=325</math>)

Latest revision as of 11:30, 26 June 2024

Problem

Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.

Solution 1

So we want $7 \nmid ab(a+b)$ and $7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2$, so we want $7^3 | a^2+ab+b^2$. Now take e.g. $a=2,b=1$ and get $7|a^2+ab+b^2$. Now by some standard methods like Hensels Lemma (used to the polynomial $x^2+x+1$, so $b$ seen as constant from now) we get also some $\overline{a}$ with $7^3 | \overline{a}^2+\overline{a}b+b^2$ and $\overline{a} \equiv a \equiv 2 \mod 7$, so $7\nmid \overline{a}b(\overline{a}+b)$ and we are done. (in this case it gives $\overline{a}=325$)

This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]

See Also

1984 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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