Every Journey Begins With One Step
Find the number of positive integer pairs ( a , b ) that satisfy the equation a 2 + a + 1 = b 2 .
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2 solutions
Nice solution! Here's my approach:
Multiply both sides by 4 and factor them gives 4 a 2 + 4 a + 4 = 4 b 2 ⇔ ( 2 b ) 2 − ( 2 a + 1 ) 2 = 3 . This tells us that the difference of two perfect squares is 3, thus 2 b = ± 2 , 2 a + 1 = ± 1 . But since a (and b ) must be a positive integer, there can't be any solution that satisfy this condition. Hence, the answer is 0.
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Nice! Your method is more suitable than mine for generalizing to the equation a 2 + a + n = b 2 , n > 1 . Mine is more of a "one-off", (pun intended), for the case n = 1 . :)
Is 0 a positive integer?
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0 is a null integer. You can tell the diference when we write the subsets of integers, for example:
• Non-negative integers: 0 + positive integers
• Positive integers: integers > 0
Assume there exists such a pair of positive integers ( a , b ) .
Since a > 0 , we see that a 2 + a + 1 > a 2 so b 2 > a 2 so b > a .
Since a and b are positive integers, there exists a positive integer k such that b = a + k . Therefore
a 2 + a + 1 = ( a + k ) 2
a 2 + a + 1 = a 2 + 2 a k + k 2
a + 1 = 2 a k + k 2
1 − k 2 = a ( 2 k − 1 )
a = 2 k − 1 1 − k 2
Clearly, for positive integers k, 1 − k 2 ≤ 0 and 2 k − 1 > 0 , hence a ≤ 0 , a contradiction. Thus there does not exist such a pair, and the answer is 0 .
Note that for a ≥ 1 we have that a 2 < a 2 + a + 1 < a 2 + 2 a + 1 = ( a + 1 ) 2 . So as a 2 + a + 1 for a ≥ 1 lies strictly between successive squares the expression cannot itself be a perfect square. Thus the number of positive integer pairs ( a , b ) is 0 .