prove that their are three different pairs of positive integer x and y. x*2 + xy + y*2 = 1729
Answers
There are more than 3 different pairs of positive integers.
They are
x = 3, y = 40
x = 8, y = 37
x = 15, y = 32
x = 23, y = 25.
Given :- prove that their are more than three different pairs of positive integer x and y. x² + xy + y² = 1729 ?
Answer :-
→ x² + xy + y² = 1729
adding xy both sides,
→ x² + xy + xy + y² = (1729 + xy)
→ x² + 2xy + y² = (1729 + xy)
→ (x + y)² = (1729 + xy)
since RHS is greater than 1729 , checking perfect square greater than 1729 we get,
→ 1681 < 1729 < 1764 .
→ (41)² < 1729 < (42)² .
now, starting from 42 we get,
- (42)² = 1764 = 1729 + xy => xy = 35 => = 7 * 5 = 35 but 7 + 5 ≠ 42 . so, No such value is possible .
- (43)² = 1849 = 1729 + xy => xy = 120 => 3 * 40 = 120 and, 3 + 40 = 43 . Therefore, x and y will be 3 and 40 .
- (44)² = 1936 = 1729 + xy => xy = 207 => = 3 * 3 * 29 => No such 2 values are possible from 3 , 3 and 29 whose sum is 44 .
- (45)² = 2025 = 1729 + xy => xy = 296 => 8 * 37 = 296 and, 8 + 37 = 45 . Therefore, x and y will be 8 and 37 .
- (46)² = 2116 = 1729 + xy => xy = 387 => = 43 * 9 and, 43 + 9 ≠ 46 . so, No such value is possible .
- (47)² = 2209 = 1729 + xy => xy = 480 => 15 * 32 = 480 and, 15 + 32 = 47 . Therefore, x and y will be 15 and 32 .
- (48)² = 2304 = 1729 + xy => xy = 575 => 23 * 25 = 575 and, 23 + 25 = 48 . Therefore, x and y will be 23 and 25 .
Therefore, Possible values of x and y will be,
- (3,40) or (40,3)
- (8,37) or (37,8)
- (15,32) or (32,15)
- (23,25) or (25,23)
Learn more :-
let a and b positive integers such that 90 less than a+b less than 99 and 0.9 less than a/b less than 0.91. Find ab/46
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