Can You Prove 2+2√(28n²+1) is a Square Number?

In summary, the "Square number challenge" is a mathematical puzzle that involves finding a specific number or set of numbers that, when squared, equal a given number. To solve this challenge, one can use various mathematical techniques such as trial and error, algebra, or a systematic approach. A square number is any number that is the result of multiplying a number by itself, and there are specific strategies that can help solve this challenge, including breaking down the given number into its prime factors, using patterns and relationships between square numbers, and using inequalities to limit the possible solutions. Furthermore, the "Square number challenge" has many real-world applications in fields such as computer science, cryptography, and engineering.
  • #1
anemone
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Let $n$ be a positive integer. Show that if $2+2\sqrt{28n^2+1}$ is an integer, then it is a square.
 
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  • #2
Let $k = 2 + 2\sqrt{28n^2 + 1}$. Then $(k-2)^2 = 4(28n^2+1)$ and so $k(k-4) = 112n^2 = 7(4n)^2.$

Since $28n^2 + 1$ is odd, so is its square root. Thus $ 2\sqrt{28n^2 + 1}$ is an odd multiple of $2$. So $k$ is an even multiple of $2$, in other words a multiple of $4$, as also is $k-4$. If $2$ occurs to the power $\alpha$ in the prime factorisation of $n$ then it occurs to the power $2\alpha+4$ in the product $k(k-4)$. But either $k$ or $k-4$ must be an odd multiple of $4$ and therefore contains $2$ to the power $2$. So the other element of that product contains $2$ to the power $2\alpha+2$. Thus the prime $2$ occurs to an even power in the prime factorisations of both $k$ and $k-4$.

If $p$ is an odd prime factor of $k$ then it is not a factor of $k-4$. If in addition $p\ne7$ then $p$ occurs to an even power in the prime factorisation of $112n^2$. So it must occur to an even power in the prime factorisation of $k$. Similarly, each prime factor of $k-4$ apart from $7$ must occur to an even power.

Finally, $7$ must be a factor of $k$ or $k-4$, but not both. So one of the numbers $k$ and $k-4$ is a square, and the other one is a multiple of $7$. If $k$ is a multiple of $7$ then $k\equiv0\pmod7$ and so $k-4\equiv3\pmod7$. But a square cannot be congruent to $3$ mod $7$. So $k$ cannot be a multiple of $7$ and is therefore a square.

It seems quite hard to find examples of positive integers $n$ such that $\sqrt{28n^2 + 1}$ is an integer. These are the only ones I could find (using the method from https://mathhelpboards.com/threads/numbers-with-a-quadratic-property.27240/#post-119314 and working with the continued fraction convergents for $\sqrt7$): $$\begin{array}{c|c|c|c} n&m = \sqrt{28n^2+1}&k = 2+2\sqrt m & \sqrt k \\ \hline 24 & 127 & 256 & 16 \\ 6096 & 32257 & 64516 & 254 \\ 1548360 & 8193151 & 16386304 & 4048. \end{array}$$
 
  • #3
Because $2\sqrt{28n^2+1} + 2$ is integer so is $\sqrt{28n^2+1}$
$\sqrt{28n^2+1}$ As it is odd let it be $2m + 1$
So $28n^2 + 1 = (2m+1)^2$
Or $28n^2 = 4m^2 + 4m$
Or $7n^2= m^2 + m = m(m+1)$
Now m and m+1 are co-primes so there exist some co-primes s and t such that $n= st$ , $m= 7s^2$ and $m+1 = t^2$ or or $m+1 = 7 s^2 $ and $m = t^2$
Now $m + 1 = 7s^2$ and $m = t^2$ => $t^2 + 1 = 7s^2$
Or $t^2 = 6 \pmod 7$
Above is not possible as $t^2 = 0/1/2/ 4 \pmod 7$
So $m + 1 = t^2$
Or $28n^2 + 1 = (2t^2-1)^2$
Or $2\sqrt{28n^2+1} + 2= 2(2t^2-1) + 2 = 4t^2 = (2t)^2$ which is a perfect square

Proved
 
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FAQ: Can You Prove 2+2√(28n²+1) is a Square Number?

What is a square number?

A square number is a number that can be formed by multiplying two equal numbers together. For example, 4 is a square number because it can be formed by multiplying 2 x 2.

How do you prove that 2+2√(28n²+1) is a square number?

To prove that 2+2√(28n²+1) is a square number, we can use the definition of a square number. We need to show that it can be written in the form of (a+b)², where a and b are integers. By expanding (a+b)², we get a² + 2ab + b². In this case, a = 1 and b = √(28n²+1). Therefore, (a+b)² = 1 + 2√(28n²+1) + (28n²+1) = (1+√(28n²+1))². Since 1+√(28n²+1) is an integer, we have proven that 2+2√(28n²+1) is a square number.

Is 2+2√(28n²+1) always a square number?

Yes, 2+2√(28n²+1) is always a square number. This is because the expression can be simplified to (1+√(28n²+1))², and the square of any integer is always a square number.

Can you provide an example of when 2+2√(28n²+1) is not a square number?

No, 2+2√(28n²+1) is always a square number. Even if n = 0, the expression simplifies to 1, which is a square number.

What is the significance of proving that 2+2√(28n²+1) is a square number?

Proving that 2+2√(28n²+1) is a square number is significant because it helps to understand the properties and relationships of different types of numbers. It also demonstrates the use of mathematical techniques and logic to solve problems and make conclusions.

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