Divisibility criterion
If the sum of two consecutive positive integers be a perfect square, then one of them must be divisible by an integral number. What is the largest value of that number?
The answer is 4.
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2 solutions
Just amazing, good on you mate!
You don't need to see anything other than 4, if it were a factor, it would have appeared just here
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Oh, do you mean in the last line? That's a good point, but it is necessary to check there's nothing larger that always divides the other member of the pair. The proof that 4 always divides a given above doesn't rule out anything about possible factors of a + 1 , or even possible other factors of a .
x + ( x + 1 ) = 2 x + 1 = A 2 ⟹
2 x = A 2 − 1 = ( A − 1 ) ( A + 1 )
since the LHS is even, so should be the RHS ( A is odd). Note that any odd A > 1 , when substituted in the equation, would give a x . Either A − 1 or A + 1 is a multiple of 4 . Now, assume a prime p > 2 is always present in either A − 1 or A + 1 . Then, we come to the contradiction that a subsequence of the sequence of even numbers, that is an arithmetic progression, is the sequence of multiples of p .
Hence, the largest positive integer that would divide x or x + 1 is 4 .
The sum of two consecutive integers is always odd; so we're looking for the square of an odd number.
That is, a + ( a + 1 ) 2 a + 1 a = ( 2 b + 1 ) 2 = 4 b 2 + 4 b + 1 = 2 b ( b + 1 )
Since one of b , b + 1 is always even, we find that a is always a multiple of 4 .
To see we can't do any better, it's enough to check the first couple of positive pairs; these are ( 4 , 5 ) and ( 1 2 , 1 3 ) , and it's easy to see that nothing larger than 4 will work.