Difference between revisions of "2021 IMO"
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For the statement to be true, there must be at least three pairs whose sum is each a perfect square. | For the statement to be true, there must be at least three pairs whose sum is each a perfect square. | ||
| − | There must be p,q,r such that p+q = x^2 and q+r = y^2 p+r = z^2. | + | There must be p,q,r such that p+q = x^2 and q+r = y^2, p+r = z^2. |
| + | |||
WLOG n<= p<= q<= r <= 2n ... Equation 1 | WLOG n<= p<= q<= r <= 2n ... Equation 1 | ||
| + | |||
p = x^2 + z^2 - y^2 | p = x^2 + z^2 - y^2 | ||
| + | |||
q = x^2 + y^2 – z^2 | q = x^2 + y^2 – z^2 | ||
| + | |||
r = y^2 + z^2 – x^2 | r = y^2 + z^2 – x^2 | ||
| + | |||
by equation 1 | by equation 1 | ||
| + | |||
2n <= x^2 + z^2 – y^2 <= 4n | 2n <= x^2 + z^2 – y^2 <= 4n | ||
| + | |||
2n <= x^2 + y^2 – z^2 <= 4n | 2n <= x^2 + y^2 – z^2 <= 4n | ||
| + | |||
2n <= y^2 + z^2 – z^2 <= 4n | 2n <= y^2 + z^2 – z^2 <= 4n | ||
| + | |||
6n <= x^2 + y^2 + z^2 <= 12n | 6n <= x^2 + y^2 + z^2 <= 12n | ||
| + | |||
6n <= 3x^2 <= 12n | 6n <= 3x^2 <= 12n | ||
| + | |||
2n <= x^2 <= 4n | 2n <= x^2 <= 4n | ||
| − | + | ||
| + | √(2n) <= x <= 2√n | ||
| + | |||
At this time n >= 100, so | At this time n >= 100, so | ||
| − | 10 * | + | 10 * √2 <= x,y,z <= 20 |
| + | |||
15 <= x,y,z <= 20 | 15 <= x,y,z <= 20 | ||
| − | where 2|x^2 + y^2 – z^2 | + | |
| + | where | ||
| + | 2|x^2 + y^2 – z^2 | ||
| + | 2|x^2 + z^2 – y^2 | ||
| + | 2|y^2 + z^2 – z^2 | ||
x = 16, y = 18, z = 20 fits perfectly | x = 16, y = 18, z = 20 fits perfectly | ||
| − | + | ||
| + | therefore the proposition is true | ||
Revision as of 11:02, 29 January 2023
For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that p+q = x^2 and q+r = y^2, p+r = z^2.
WLOG n<= p<= q<= r <= 2n ... Equation 1
p = x^2 + z^2 - y^2
q = x^2 + y^2 – z^2
r = y^2 + z^2 – x^2
by equation 1
2n <= x^2 + z^2 – y^2 <= 4n
2n <= x^2 + y^2 – z^2 <= 4n
2n <= y^2 + z^2 – z^2 <= 4n
6n <= x^2 + y^2 + z^2 <= 12n
6n <= 3x^2 <= 12n
2n <= x^2 <= 4n
√(2n) <= x <= 2√n
At this time n >= 100, so
10 * √2 <= x,y,z <= 20
15 <= x,y,z <= 20
where
2|x^2 + y^2 – z^2 2|x^2 + z^2 – y^2 2|y^2 + z^2 – z^2 x = 16, y = 18, z = 20 fits perfectly
therefore the proposition is true
