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- Feb 14, 2012

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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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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,642

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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- Feb 7, 2012

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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.

- Mar 31, 2013

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$\sqrt{28n^2+1}$ 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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