2004 JBMO Problems/Problem 3: Difference between revisions
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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 | ||
-igetit | |||
Latest revision as of 19:09, 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
-igetit
