Difference between revisions of "2017 USAJMO Problems/Problem 4"
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==Problem== | ==Problem== | ||
− | Are there any triples <math>(a,b,c)</math> of positive integers such that <math>(a-2)(b-2)(c-2) + 12</math> is prime that properly divides the number <math>a^2 + b^2 + c^2 + abc - 2017</math>? | + | Are there any triples <math>(a,b,c)</math> of positive integers such that <math>(a-2)(b-2)(c-2) + 12</math> is prime that properly divides the positive number <math>a^2 + b^2 + c^2 + abc - 2017</math>? |
− | ==Solution== | + | ==Solution 1== |
− | + | The answer is no. Substitute <math>x=a-2,y=b-2,z=c-2</math>. This means that <math>x,y,z\geq -1</math>. Then <cmath>a^2+b^2+c^2+abc-2017=(x+y+z-41)(x+y+z+49)+xyz+12.</cmath> It is given in the problem that this is positive. Now, suppose for the sake of contradiction that <math>xyz+12</math> is a prime. Clearly <math>x,y,z\neq 0</math>. Then we have <cmath>\frac{(x+y+z-41)(x+y+z+49)}{xyz+12}</cmath> is an integer greater than or equal to <math>1</math>. This also implies that <math>x+y+z > 41</math>. Since <math>xyz+12</math> is prime, we must have <cmath>xyz+12\mid x+y+z-41\text{ or } xyz+12\mid x+y+z+49.</cmath> Additionally, <math>x, y, z</math> must be odd, so that <math>xyz+12</math> is odd while <math>x+y+z-41,x+y+z+49</math> are even. So, if <cmath>xyz+12\mid x+y+z-41\text{ or }xyz+12\mid x+y+z+49,</cmath> we must have <cmath>2(xyz+12)\leq x+y+z-41\text{ or }2(xyz+12)\leq x+y+z+49.</cmath> Now suppose WLOG that <math>x=-1</math> and <math>y,z>0</math>. Then we must have <math>yz\leq 10</math>, impossible since <math>x+y+z>41</math>. Again, suppose that <math>x,y=-1</math> and <math>z>0</math>. Then we must have <cmath>2(z+12)\leq z-43\text{ or }2(z+12)\leq z+47,</cmath> and since in this case we must have <math>z>43</math>, this is also impossible. | |
− | {{ | + | Then the final case is when <math>x,y,z</math> are positive odd numbers. Note that if <math>xyz>x+y+z</math> for positive integers <math>x,y,z</math>, then <math>abc>a+b+c</math> for positive integers <math>a,b,c</math> where <math>a>x,b>y,c>z</math>. Then we only need to prove the case where <math>x+y+z=43</math>, since <math>x+y+z</math> is odd. Then one of <cmath>2(xyz+12)\leq 2\text{ and/or }2(xyz+12)\leq 92</cmath> is true, implying that <math>xyz\leq -11</math> or <math>xyz\leq 34</math>. But if <math>x+y+z=43</math>, then <math>xyz</math> is minimized when <math>x=1,y=1,z=41</math>, so that <math>xyz\geq 41</math>. This is a contradiction, so we are done. |
==See also== | ==See also== | ||
{{USAJMO newbox|year=2017|num-b=3|num-a=5}} | {{USAJMO newbox|year=2017|num-b=3|num-a=5}} |
Latest revision as of 12:44, 3 April 2024
Problem
Are there any triples of positive integers such that
is prime that properly divides the positive number
?
Solution 1
The answer is no. Substitute . This means that
. Then
It is given in the problem that this is positive. Now, suppose for the sake of contradiction that
is a prime. Clearly
. Then we have
is an integer greater than or equal to
. This also implies that
. Since
is prime, we must have
Additionally,
must be odd, so that
is odd while
are even. So, if
we must have
Now suppose WLOG that
and
. Then we must have
, impossible since
. Again, suppose that
and
. Then we must have
and since in this case we must have
, this is also impossible.
Then the final case is when
are positive odd numbers. Note that if
for positive integers
, then
for positive integers
where
. Then we only need to prove the case where
, since
is odd. Then one of
is true, implying that
or
. But if
, then
is minimized when
, so that
. This is a contradiction, so we are done.
See also
2017 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |