Difference between revisions of "2004 JBMO Problems/Problem 3"
(Created page with "Let 4x + 3y = a^2 3x + 4y = b^2 Then 7(x + y) = a^2 + b^2 But any perfect square can only be congruent to 0,1,2, or 4 modulo 7 Thus a = 7p b = 7q x + y = 7(p^2 + q^...") |
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− | Let 4x + 3y = a^2 | + | Let |
− | + | 4x + 3y = a^2 | |
+ | 3x + 4y = b^2 | ||
Then 7(x + y) = a^2 + b^2 | Then 7(x + y) = a^2 + b^2 | ||
+ | |||
But any perfect square can only be congruent to 0,1,2, or 4 modulo 7 | But any perfect square can only be congruent to 0,1,2, or 4 modulo 7 | ||
− | Thus a = 7p | + | Thus |
− | + | ||
+ | a = 7p | ||
+ | |||
+ | b = 7q | ||
+ | |||
x + y = 7(p^2 + q^2) | x + y = 7(p^2 + q^2) | ||
+ | |||
x + y is congruent to 0 mod 7. | x + y is congruent to 0 mod 7. | ||
4x + 3y = a^2 = 49p^2 | 4x + 3y = a^2 = 49p^2 | ||
+ | |||
3(x+y) + x = 49p^2 | 3(x+y) + x = 49p^2 | ||
+ | |||
x is congruent to 0 mod 7 | x is congruent to 0 mod 7 | ||
Similarly, we get y is congruent to 0 mod 7 | Similarly, we get y is congruent to 0 mod 7 |
Revision as of 19:08, 10 April 2019
Let 4x + 3y = a^2 3x + 4y = b^2
Then 7(x + y) = a^2 + b^2
But any perfect square can only be congruent to 0,1,2, or 4 modulo 7
Thus
a = 7p
b = 7q
x + y = 7(p^2 + q^2)
x + y is congruent to 0 mod 7.
4x + 3y = a^2 = 49p^2
3(x+y) + x = 49p^2
x is congruent to 0 mod 7
Similarly, we get y is congruent to 0 mod 7