So, the LHS is: $$ \begin{align} a^2 + b^2 &= 4(m^2+n^2) + 4(m+n) + 2 \\\\ &= 2\left( 2(m^2+n^2) + 2(m+n)+1 \right) \\\\ &= 2\times\text{an odd number} \end{align} $$ This can not be a perfect square, as it doesn't have an even power of $2$.
Another way to find number of inversions in such an array, is to notice that there will be. n − 1. n - 1 n −1 inversions with the first index as the first element of the inversion pair, n − 2. n -2 n−2 inversions with the second index as the first element of the inversion pair, and so on. And finally, zero inversions with the.
Answer (1 of 5): A2A, thanks. If a and b were both odd, then c^2 would be even, hence so is c. Now write a = 2m+1, b = 2n+1, c = 2k, where m, n, k are integers. It follows that 4 k^2 = {\bf c^2 = a^2 + b^2} = 4m^2 + 4m + 1 + 4n^2 + 4n + 1. Dividing both sides by 4 gives k^2 = (a non-integer...
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If both and are odd, then their squares will also be odd and hence the sum will be even. This will demand from to be even and therefore will be even. Now since is even, then there exists positive integer such that:
However, since and are integer, can’t be an integer. Thus, can’t be an integer, since any integer squared is itself an integer. However, we have previously derived that itself is an integer.
Note that the square of an odd integer is odd itself (proof an odd integer has the form its square is ) and an odd integer plus an other odd integer makes an even one () so it would be a contradiction that all are odd. Now to proof that the even integer can’t be (if the others are odd):
The positive integers and both end in the same sequence of four digits when written in base , where digit a is not zero. Find the three-digit number .
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This can happen for only 3 values : 1, 5 and 6.
It can also be proven directly from a certain formula. The numbers a, b and c such that a 2 + b 2 = c 2 are called "Pythagorean triples." Given two integers m and n satisfying 0 < m < n, you can obtain a Pythagorean triple with this "recipe":
We know , a,b,c are integers. Now some basic principles:
So a and c could be odd, but b is even, and that's enough for a b c to be even as well. For example: 5 2 + 12 2 = 13 2 and 5 × 12 × 13 = 780.
The product of three consecutive positive integers is times their sum. What is the sum of their squares?
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The arithmetic mean of the nine numbers in the set is a -digit number , all of whose digits are distinct. The number does not contain the digit. Solution.
Squaring a number is just multiplying it by itself, so the laws of modulo arithmetic should all apply. So if we’re dealing with modulo four, there are only four kinds of numbers. Also 0^2 = 0*0 = 0 mod 4, 1^2 = 1*1 = 1 mod 4, 2^2 = 2*2 = 0 mod 4, and 3^2 = 3*3 = 1 mod 4. So any square is either 0 or 1 mod 4. To prove the statement by contradiction, I assume that it is possible to find integers a, b, and c such that a^2 + b^2 = c^2 and a and b are both odd. That means that a^2 and b^2 are each 1 mod 4, since if they’re 0 mod 4 they wouldn’t be odd. If they’re each 1 mod 4, then their sum is 2 m
Now, since a 2 + b 2 = c 2, ( 1) and ( 2) are incompatible and this is our contradiction.
But a 2 + b 2 is the sum of two odd squares, therefore, as we have seen, it is not a multiple of 4.
So b is even since square of even number is even.
So the sum of two odd squares is a multiple of 4 plus 2, which implies that it is even but not a multiple of 4.