All perfect squares
a 2 + b + c b 2 + c + a c 2 + a + b
If a , b , c are positive integers, then is it possible that all three of the numbers above are perfect squares?
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2 solutions
Why can't we say that the maximum number of squares is 0 or 1?
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Because an example was already provided which had 2 perfect squares (1^2 + 1 + 7 = 9, 1^2 + 7 + 1 = 9).
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What if we don't think about examples as it's a toiling work ? Is there a mathematical way to solve such questions?
Thank you!
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@Kaushik Chandra – The shortcut way here is to notice symmetry.
a^2 + b + c
b^2 + a + c
Suppose that a = b (which is allowed by the rules of the problem)
a^2 + a + c
a^2 + a + c
That means these add to the same sum! We can set a to any positive integer we like, and then just find some value of c that goes along with it that makes both sums to be perfect squares.
Why is a 2 + b + c ≥ ( a + 1 ) 2 ?
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Because a 2 + b + c is a perfect square and bigger, than a 2 , so the minimum value of it, is the next perfect square (after a 2 ), which is ( a + 1 ) 2 .
You meant a, b and c are strictly positive. Otherwise, you can choose a=1; b=0 and c=0. In this case, all the equations are perfect square.
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Zero is neither positive nor negative, so by saying positive integer, you eliminate zero as an option.
Sorry all the numbers are perfect square not equations.
Given a = b = c = 1, all 3 expressions evaluate to 4. 4 being 2², it is indeed possible for all 3 expressions to yield prefect squares. No restriction was given that a, b, and c must be distinct.
It is clear that two of them can be perfect squares by taking ( a , b , c ) = ( 1 , 1 , 2 ) .
Assume that there are positive integers a , b , c such that all the three numbers are perfect squares. Without loss of generality, assume a ≤ b ≤ c . Then, we must have
c 2 < c 2 + a + b ≤ c 2 + c + c < ( c + 1 ) 2
This implies that the third number is bounded between two consecutive perfect squares, which is not possible to be a perfect square.
Hence, the maximum number of perfect squares among the three numbers is 2.
WOah, nice bounding. I didn't expect such a simple inequality makes this question so simple to tackle. Thanks for sharing!
Simple and elegant. Nice.
it is also possible with (a,b,c) = (6,6,7)
It is possible two of the numbers are perfect squares. Here is an example: ⎩ ⎪ ⎨ ⎪ ⎧ a = 1 b = 1 c = 7
Suppose it is possible three of the numbers are perfect squares. Since a , b , c are three positive integers,
a 2 + b + c ⇒ b + c b 2 + a + c ⇒ a + c c 2 + a + b ⇒ a + b ≥ ( a + 1 ) 2 = a 2 + 2 a + 1 ≥ 2 a + 1 ≥ ( b + 1 ) 2 = b 2 + 2 b + 1 ≥ 2 b + 1 ≥ ( c + 1 ) 2 = c 2 + 2 c + 1 ≥ 2 c + 1
By adding the equations, we get: 2 a + 2 b + 2 c ≥ 2 a + 2 b + 2 c + 3 which is a contradiction.
So the answer is 2 .