How many pairs of integers (not necessarily positive) are there such that both a2+6b2 and b2+6a2 are both squares?
There exists only one solution and the whole challenge is in proving the fact that there does not exist any other solution. So here we go...
Let's add both these expressions. Then as per the problem,
7(a^2+b^2)=c^2+d^2
Clearly LHS is a multiple of 7 and so RHS must be a multiple of 7 in order for the solution to exist. Now, let's investigate the RHS.
Both c and d can be expressed as 7k, 7k+1, 7k+2,7k+3,7k+4, 7k+5,7k+6.
So, remainder when (c^2+d^2) is divided by 7 can be the same if one among the following is divided by7:
0, (1+2^2), (1+3^2), (1+4^2), (1+5^2), (1+6^2)
(2^2+3^2), (2^2+4^2), (2^2+5^2), (2^2+6^2)
(3^2+4^2), (3^2+5^2), (3^2+6^2)
(4^2+5^2), (4^2+6^2)
(5^2+6^2)
None of the above except 0 is divisible by 7.
Hence there exists only one solution for (a,b) -> (0,0)
This is a problem posted on Brilliant(dot)org
The above solution is wrong / incomplete.
-Calvin Lin
Brilliant Challenge Master
To find the number of pairs of integers (a, b) such that both a^2 + 6b^2 and b^2 + 6a^2 are squares, we can use a systematic approach.
Let's analyze the possible values of a^2 + 6b^2 and b^2 + 6a^2 individually and then find the common solutions.
1. For a^2 + 6b^2 to be a square, a^2 must be a multiple of 6, and b^2 must be a multiple of both 1 and 6. Hence, a must be a multiple of 2 and b must be a multiple of 1.
2. For b^2 + 6a^2 to be a square, b^2 must be a multiple of 6, and a^2 must be a multiple of both 1 and 6. Hence, b must be a multiple of 2 and a must be a multiple of 1.
Considering both conditions together, we conclude that a must be a multiple of 2, and b must be a multiple of 2.
Let's consider the cases:
Case 1: a and b are both even
If a and b are both even, let's suppose a = 2x and b = 2y, where x and y are integers.
Now we need to find the number of pairs (x, y) such that (2x)^2 + 6(2y)^2 and (2y)^2 + 6(2x)^2 are both squares.
Simplifying, we get 4x^2 + 24y^2 and 4y^2 + 24x^2.
2x^2 + 12y^2 and 2y^2 + 12x^2 are both squares.
We can simplify further by dividing everything by 2 to get x^2 + 6y^2 and y^2 + 6x^2.
Now we have reduced the problem to finding the number of pairs (x, y) such that x^2 + 6y^2 and y^2 + 6x^2 are both squares.
Case 2: a and b are both odd
If a and b are both odd, let's suppose a = 2x + 1 and b = 2y + 1, where x and y are integers.
Now we need to find the number of pairs (x, y) such that (2x + 1)^2 + 6(2y + 1)^2 and (2y + 1)^2 + 6(2x + 1)^2 are both squares.
Expanding, we get 4x^2 + 4x + 1 + 24y^2 + 24y + 6 and 4y^2 + 4y + 1 + 24x^2 + 24x + 6.
Simplifying, we have 4x^2 + 24y^2 + 4x + 24y + 7 and 4y^2 + 24x^2 + 4y + 24x + 7.
We can simplify further by excluding the constants to get x^2 + 6y^2 + x + 6y and y^2 + 6x^2 + y + 6x.
Again, we have reduced the problem to finding the number of pairs (x, y) such that x^2 + 6y^2 + x + 6y and y^2 + 6x^2 + y + 6x are both squares.
In both cases, we need to find the number of pairs (x, y) such that x^2 + 6y^2 and y^2 + 6x^2 are both squares.
To analyze further and find the exact number of pairs, we need to examine the properties of numbers that satisfy these conditions.
Note: There are infinitely many positive and negative integer solutions to these equations.