- Thread starter
- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$m,n \in N$

satisfying :

$\sqrt {m+2007}+\sqrt {m-325}=n $

find Max(n)

satisfying :

$\sqrt {m+2007}+\sqrt {m-325}=n $

find Max(n)

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

$m,n \in N$

satisfying :

$\sqrt {m+2007}+\sqrt {m-325}=n $

find Max(n)

satisfying :

$\sqrt {m+2007}+\sqrt {m-325}=n $

find Max(n)

- Jan 26, 2012

- 644

$$\sqrt{u + 2332} + \sqrt{u} = n$$

For $n$ to be an integer, both square roots have to be integers*. So we are looking for two (positive) squares which differ by $2332$ units. That is, we need to solve:

$$x^2 - y^2 = 2332$$

Doing some factoring:

$$(x - y)(x + y) = 2^2 \times 11 \times 53$$

After some observation** we have the following solution pairs:

$$(x, y) \to \{ ( 64, 42), ~ (584, 582) \}$$

It follows that the only solutions for $u$ are $u = 42^2$ and $u = 582^2$. This gives $n = 106$ and $n = 1166$ (and $m = 2089$, $m = 339049$ respectively).

Therefore $\max{(n)} = 1166$. $\blacksquare$

* perhaps this part merits justification, you can check Square roots have no unexpected linear relationships | Annoying Precision which is pretty hardcore (perhaps a reduced argument suffices here) but I will take it that the question assumed this to be true. Other approaches which do not rely on this assumption would be interesting to see as well.

** this can be solved efficiently by using every reasonable factor combination for $x - y$ and deducing $x + y$ from it.[/JUSTIFY]

Last edited:

- Moderator
- #3

- Feb 7, 2012

- 2,793

Last edited:

- Mar 31, 2013

- 1,346

we have

(m+ 2007) – (m- 325) = 2332 ..1

√(m+2007)+√(m−325) = n ..(2) given

Dividing (2) by (1) we get

√(m+2007)-√(m−325) = 2332/n … (3)

And (2) and (3) to get 2 √(m+2007) = n + 2332/n

Now RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.

- Thread starter
- #5

- Jan 25, 2013

- 1,225

$u^2=m-325-------(1)$

$(u+a)^2=m+2007-------(2)$

$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)

max(n) must happen while a=2,so from (3) we have u=582

so max(n)=582+584=1166

- Moderator
- #6

- Feb 7, 2012

- 2,793

Very neat!we have

(m+ 2007) – (m- 325) = 2332 ..1

√(m+2007)+√(m−325) = n ..(2) given

Dividing (2) by (1) we get

√(m+2007)-√(m−325) = 2332/n … (3)

And (2) and (3) to get 2 √(m+2007) = n + 2332/n

Now RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.

- Thread starter
- #7

- Jan 25, 2013

- 1,225

now ,find min(n)=?

- Mar 31, 2013

- 1,346

√(m+2007)+√(m−325) = nnow ,find min(n)=?

put m- 325 = p (which is >=0)

we get √(p+2332)+√p which is >= sqrt(2332) >= 49

now we should look for lowest even factor of 2332(even : reason same as my previous solution) which is >= 49

now 2332 = 2 * 2 * 11 * 53 and lowest even factor of 2332 >= 49 is 2 * 53 = 106

- Thread starter
- #9

- Jan 25, 2013

- 1,225

$2332=a^2+2ua=a(2u+a)=a\times n=2\times 1166=22\times 2\times53=22\times 106$

$u^2=m-325-------(1)$

$(u+a)^2=m+2007-------(2)$

$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)

max(n) must happen while a=2,so from (3) we have u=582

so max(n)=582+584=1166

so max(n)=1166 ,and a=2

min(n)=106, and a=22

(here a,and n=2u+a are all even numbers)