For any k in the natural numbers, define n = 2k² + 1. then (n-1)² + (2k)², n² + 0², n² + 1² fulfill the desired property, as (n-1)² + (2k)² = n² + 2n + 1 + 4k² = n² - 2(2k² + 1) + 4k² + 1 = n² - 1 The trick was to choose 2n - 2 as a square number.
Yeah did (almost) the same trick: building a solution around the term (2k² + 1)². So we get: n = (2k² + 1)² - 1 = (2k²)² + 2*2k² + 1 - 1= (2k²)² + (2k)² n + 1 = (2k² + 1)² + 0² n + 2 = (2k² + 1)² + 1²
My solution was to start with the equation: (n+1)^2 = n^2 + 2n + 1 Now, consider any integer n such that 2n is a square number. This will happen whenever n = 2k^2 for some integer k (so: n = 0, 2, 8, 18, 32,...). Thus, there are infinitely many such n. Now, for any n satisfying the above, consider N = (n+1)^2 - 1. We will then have: N = (n+1)^2 - 1 N = n^2 + 2n + 1 - 1 N = n^2 + 2(2k^2) N = n^2 + (2k)^2 N+1 = (n+1)^2 - 1 + 1 N+1 = (n+1)^2 + 0^2 N+2 = (n+1)^2 - 1 + 2 N+2 = (n+1)^2 + 1^2 Thus (N,N+1,N+2) forms a triple satisfying the required condition. The first five examples of such triples are: 0 = 0^2 + 0^2, 1 = 1^2 + 0^2, 2 = 0^2 + 1^2 8 = 2^2 + 2^2, 9 = 3^2 + 0^2, 10 = 3^2 + 1^2 80 = 8^2 + 4^2, 81 = 9^2 + 0^2, 82 = 9^2 + 1^2 360 = 18^2 + 6^2, 361 = 19^2 + 0^2, 362 = 19^2 + 1^2 1088 = 32^2 + 8^2, 1089 = 33^2 + 0^2, 1090 = 33^2 + 1^2
We can clean this proof up a bit. Let n=(2k^2+1)^2-1. Then as noted, n+1 = (2k^2+1)^2 + 0^2 and n+2 is (2k^2+1)^2+1^2. n=(2k^2+1)^2-1 = 4k^4 + 4k^2 = (2k^2)^2 + (2k)^2 as required. This is a "magic" version of the proof I'd admit. A clean non-magic would be "Let n=a^2-1. Then n+1 = a^2+0^2 and n+2 is a^2+1^2. If a=x+1, then n=x^2+2x; the -1 cancels out. If 2x is a square, we are done. And 2x is a square when x=2k^2 for some k." Which is basically your proof, just jot flipped upside down.
Here is my solution: First notice n^2, n^2 + 1^2 already form 2 consecutive numbers that are sum of 2 square numbers. We want to make n^2+2 also a sum of 2 square numbers. To do this, consider a special case where n^2+2 = (n-1)^2 + m^2 (Inspired by 16, 17, 18 triplet) Then we will have 2n = m^2-1 = (m+1)(m-1) Since left side is even, the right side has to be even. But if one of the factor on the right is even, then both factors are even. So we can rewrite n = 2p, m+1=2q, and we have 4p = (2q)(2q-2), or equivalently, p=q(q-1) i.e. If p is a product of 2 consecutive positive integers, then we have the desired results. Algebraically, we have: [2q(q-1)]^2 + 0^2 = 4q^4-8q^3+4q^2 [2q(q-1)]^2 + 1^2 = 4q^4-8q^3+4q^2 + 1 [2q(q-1)-1]^2 + (2q-1)^2 = 4q^4-8q^3+4q^2 + 2 Examples are: 16=4^2+0^2, 17=4^2+1^2, 18=3^2+3^2 144=12^2+0^2, 145=12^2+1^2, 146=11^2+5^2
Let n+1=m^2, then n+2=m^2+1^2. Just need to choose m such that n=a^2+b^2. m^2-1=a^2+b^2 => (m+a)(m-a)=b^2+1. Choose b=10k+2. Since 5 | b^2+1, we can let m-a=5, and m = ((b^2+1)/5+5)/2.
For any k in the natural numbers, define n = 2k² + 1.
then (n-1)² + (2k)², n² + 0², n² + 1² fulfill the desired property, as
(n-1)² + (2k)² = n² + 2n + 1 + 4k² = n² - 2(2k² + 1) + 4k² + 1 = n² - 1
The trick was to choose 2n - 2 as a square number.
Yeah did (almost) the same trick: building a solution around the term (2k² + 1)².
So we get:
n = (2k² + 1)² - 1 = (2k²)² + 2*2k² + 1 - 1= (2k²)² + (2k)²
n + 1 = (2k² + 1)² + 0²
n + 2 = (2k² + 1)² + 1²
@@MrGeorge1896 Cool, we think alike :-)
lowkey asmr
My solution was to start with the equation:
(n+1)^2 = n^2 + 2n + 1
Now, consider any integer n such that 2n is a square number. This will happen whenever n = 2k^2 for some integer k (so: n = 0, 2, 8, 18, 32,...). Thus, there are infinitely many such n.
Now, for any n satisfying the above, consider N = (n+1)^2 - 1. We will then have:
N = (n+1)^2 - 1
N = n^2 + 2n + 1 - 1
N = n^2 + 2(2k^2)
N = n^2 + (2k)^2
N+1 = (n+1)^2 - 1 + 1
N+1 = (n+1)^2 + 0^2
N+2 = (n+1)^2 - 1 + 2
N+2 = (n+1)^2 + 1^2
Thus (N,N+1,N+2) forms a triple satisfying the required condition.
The first five examples of such triples are:
0 = 0^2 + 0^2, 1 = 1^2 + 0^2, 2 = 0^2 + 1^2
8 = 2^2 + 2^2, 9 = 3^2 + 0^2, 10 = 3^2 + 1^2
80 = 8^2 + 4^2, 81 = 9^2 + 0^2, 82 = 9^2 + 1^2
360 = 18^2 + 6^2, 361 = 19^2 + 0^2, 362 = 19^2 + 1^2
1088 = 32^2 + 8^2, 1089 = 33^2 + 0^2, 1090 = 33^2 + 1^2
Crazy. Thanks!
@@idk7016 No problem :) !
We can clean this proof up a bit. Let n=(2k^2+1)^2-1. Then as noted, n+1 = (2k^2+1)^2 + 0^2 and n+2 is (2k^2+1)^2+1^2.
n=(2k^2+1)^2-1 = 4k^4 + 4k^2 = (2k^2)^2 + (2k)^2 as required.
This is a "magic" version of the proof I'd admit.
A clean non-magic would be "Let n=a^2-1. Then n+1 = a^2+0^2 and n+2 is a^2+1^2.
If a=x+1, then n=x^2+2x; the -1 cancels out. If 2x is a square, we are done. And 2x is a square when x=2k^2 for some k."
Which is basically your proof, just jot flipped upside down.
Here is my solution:
First notice n^2, n^2 + 1^2 already form 2 consecutive numbers that are sum of 2 square numbers. We want to make n^2+2 also a sum of 2 square numbers.
To do this, consider a special case where n^2+2 = (n-1)^2 + m^2
(Inspired by 16, 17, 18 triplet)
Then we will have 2n = m^2-1 = (m+1)(m-1)
Since left side is even, the right side has to be even.
But if one of the factor on the right is even, then both factors are even.
So we can rewrite n = 2p, m+1=2q, and we have
4p = (2q)(2q-2), or equivalently, p=q(q-1)
i.e. If p is a product of 2 consecutive positive integers, then we have the desired results.
Algebraically, we have:
[2q(q-1)]^2 + 0^2 = 4q^4-8q^3+4q^2
[2q(q-1)]^2 + 1^2 = 4q^4-8q^3+4q^2 + 1
[2q(q-1)-1]^2 + (2q-1)^2 = 4q^4-8q^3+4q^2 + 2
Examples are:
16=4^2+0^2, 17=4^2+1^2, 18=3^2+3^2
144=12^2+0^2, 145=12^2+1^2, 146=11^2+5^2
interesting
Let n+1=m^2, then n+2=m^2+1^2. Just need to choose m such that n=a^2+b^2.
m^2-1=a^2+b^2 => (m+a)(m-a)=b^2+1. Choose b=10k+2. Since 5 | b^2+1, we can let m-a=5, and m = ((b^2+1)/5+5)/2.
1 is a square. All natural numbers except 1 can be defined as n+1. Therefore all sequences for n>1, n is whole