In honour of 1729
Just as 1729 is the smallest positive integer that can be expressed as the sum of 2 distinct positive cubes in 2 different ways, what is the smallest positive integer that can be expressed as the sum of 2 distinct positive squares in 2 different ways?
The answer is 65.
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2 solutions
Moderator note:
The logic in this solution is incorrect. In particular, it is not necessarily true that "To minimize N, we let one of the smaller factors be 1". There is no reason why if one of the smaller factors is 2, we can't get a smaller result. In fact, we do.
In the above solution, there is no mention / use of the fact that a = b and c = d , and hence this does not yet preclude the solution of N = 5 0 with ( a , b ) = ( 1 , 7 ) , ( c , d ) = ( 5 , 5 ) .
This problem is nice one ... you must have given it a topic and a level ..... @Calvin Lin
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Thank you for your encouragement! Can we be friends?
Why isn't the answer 5 0 = 1 2 + 7 2 = 5 2 + 5 2 ?
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Also , its not mentioned 'distinct' positive integers...
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Only the pairs are distinct. Within a pair, the numbers do not need to be distinct. IE we have ( a , b ) = ( 1 , 7 ) , ( c , d ) = ( 5 , 5 ) which satisfies the condition of b − d > 0 .
Note that the initial part or the argument is false, and there is no reason why the smaller factor must be 1. Instead, by letting it be 2, we can minimize the total instead.
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@Calvin Lin – Ok! got it !
@Calvin Lin – Dear Calvin,
I already amended the problem to say that even within the pair, the squares have to be distinct.
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@Noel Lo – Sorry about the error, it was late for me.
As mentioned, your solution is flawed. Can you provide a different solution to show that 65 is indeed the minimum?
5 squared + 6 squared is NOT 50.
It is not true that "to minimize N, we let one of the smaller factors be 1".
Let the smallest integer be N = a 2 + b 2 = c 2 + d 2 .Now a 2 - c 2 = d 2 - b 2 or (a+c)(a-c) = (d+b)(d-b). To minimise N, we let one of the smaller factors (i.e. a-c or d-b) be 1. WLOG, we let a-c=1. This indicates that a and c are of different parity (i.e. one odd one even) hence a+c is also odd.
In other words, both b+d and d-b are odd too. Note that d-b > 1 in order to have two distinct pairs of squares. Moreover, d+b cannot be equal to d-b otherwise b=0 but all squares need to be positive as mentioned. So the smallest possible value for d-b is 3 while that of d+b is 5 since both need to be odd.
It follows that a=8, b=1, c=7 and d=4. So we have N = 8 2 + 1 2 = 7 2 + 4 2 = 6 5 .