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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)
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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>)
  
 
This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [https://aops.com/community/p366644]
 
This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [https://aops.com/community/p366644]
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 +
== Solution 2 ==
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=== Lemma: === <math>a^{7} \equiv a (mod 7) </math>.
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=== Proof: === Recall that if <math>7 \nmid x </math>, then <math>x^{3} \equiv 1, -1 (mod 7) </math>. Therefore, <math>x^{6} \equiv 1 (mod 7) </math>.
 +
<math>\implies a^{7} \equiv a (mod 7) \forall a \nequiv 0 (mod 7) </math>.
 +
 +
However, if <math>7|x </math>, then <math>a^{7} \equiv a (mod 7) </math>.
 +
 +
So now, we need to find one pair of integers (a, b) such that <math>(a+b)^{7} - a^{7} - b^{7} \equiv 0 (mod 7) </math>. This means <math>(a+b)^{7} \equiv a^{7} + b^{7} (mod 7) </math>.
 +
<math>\implies (a+b) \equiv a+b (mod 7) </math>. But this is true for all pairs of integers (a, b). So any random pair of integers would work.
 +
 +
Footnote: Even a pair of integers (a, b) which satisfies <math>7|ab(a+b) </math> would work. So the condition given is irrelevant. Try it!
  
 
== See Also == {{IMO box|year=1984|num-b=1|num-a=3}}
 
== See Also == {{IMO box|year=1984|num-b=1|num-a=3}}

Revision as of 08:00, 25 May 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]

Solution 2

=== Lemma: === $a^{7} \equiv a (mod 7)$. === Proof: === Recall that if $7 \nmid x$, then $x^{3} \equiv 1, -1 (mod 7)$. Therefore, $x^{6} \equiv 1 (mod 7)$. $\implies a^{7} \equiv a (mod 7) \forall a \nequiv 0 (mod 7)$ (Error compiling LaTeX. Unknown error_msg).

However, if $7|x$, then $a^{7} \equiv a (mod 7)$.

So now, we need to find one pair of integers (a, b) such that $(a+b)^{7} - a^{7} - b^{7} \equiv 0 (mod 7)$. This means $(a+b)^{7} \equiv a^{7} + b^{7} (mod 7)$. $\implies (a+b) \equiv a+b (mod 7)$. But this is true for all pairs of integers (a, b). So any random pair of integers would work.

Footnote: Even a pair of integers (a, b) which satisfies $7|ab(a+b)$ would work. So the condition given is irrelevant. Try it!

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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