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Root equation

Albert

Well-known member
Jan 25, 2013
1,225
$m,n \in N$

satisfying :

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

find Max(n)
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Re: root equation

[JUSTIFY]Apply the variable substitution $u = m - 325$, and hence $m + 2007 = u + 2332$. We get:
$$\sqrt{u + 2332} + \sqrt{u} = n$$
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]
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: root equation

As a slight variation on Bacterius's solution, take $u$ to be $m+841$ rather than $m-325$. The equation then becomes $$\sqrt{u+1166} + \sqrt{u-1166} = n.$$ Square both sides to get $$2u + 2\sqrt{u^2-1166^2} = n^2.$$ For an integer solution, $u^2-1166^2$ must be a square, say $u^2-1166^2 = w^2.$ Thus $(w,1166,u)$ is a pythagorean triple, which must be of the form $(p^2-q^2,\,2pq,\,p^2+q^2)$ for some integers $p,q$ with $p>q$. In particular, $1166=2pq$. But $1166 = 2*11*53$, so the only possibilities are $p=53,\ q=11$, or $p=583,\ q=1.$ That then leads to the same solutions as those of Bacterius.
 
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kaliprasad

Well-known member
Mar 31, 2013
1,309
Re: root equation

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.
 

Albert

Well-known member
Jan 25, 2013
1,225
let :
$u^2=m-325-------(1)$
$(u+a)^2=m+2007-------(2)$
$(2)-(1):2332=a^2+2ua-----(3)$ (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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: root equation

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.
Very neat! (Clapping)
 

Albert

Well-known member
Jan 25, 2013
1,225
now ,find min(n)=?
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
now ,find min(n)=?
√(m+2007)+√(m−325) = 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
 

Albert

Well-known member
Jan 25, 2013
1,225
let :
$u^2=m-325-------(1)$
$(u+a)^2=m+2007-------(2)$
$(2)-(1):2332=a^2+2ua-----(3)$ (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
$2332=a^2+2ua=a(2u+a)=a\times n=2\times 1166=22\times 2\times53=22\times 106$
so max(n)=1166 ,and a=2
min(n)=106, and a=22
(here a,and n=2u+a are all even numbers)