## y**2 + 5 x y + 5 x**2

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• ... Thank you. Now it is clear that 5 is indeed a quadratic residue of p = (137^137 + 1992)*(137^137 + 3464) And therefore, once we find the sqrt(5) mod p. we
Message 1 of 2 , Oct 4, 2010
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> ________________________________________________________________________
> 1c. Re: qfbsolve exercise
> Date: Sun Oct 3, 2010 4:24 pm ((PDT))

> Kermit Rose<kermit@...> wrote:
>
>>> Exercise: Find two pairs of positive integers (x,y) such that
>>> 5*x^2 + 5*x*y + y^2 = (137^137 + 1992)*(137^137 + 3464)
>
>> (137^137 + 1992)*(137^137 + 3464) = 1 * 3 = 3 mod 5.
>

> 137 = 2 mod 5
> 137 = 1 mod 4
> 137^137 = 2^1 mod 5
> 137^137 + 1992 = 2 + 2 mod 5 = 4 mod 5
> 137^137 + 3464 = 2 + 4 mod 5 = 1 mod 5
>
> David

Thank you.

Now it is clear that 5 is indeed a quadratic residue of
p = (137^137 + 1992)*(137^137 + 3464)

And therefore, once we find the sqrt(5) mod p.

we need only find the x such that

if y = [(5 + sqrt(5))/2 ] x mod p

or y = [(5 - sqrt(5))/2 ] x mod p,

then

( y + x [(5 + sqrt(5) )/2 ] ) ( y + x [ (5 - sqrt(5))/2 ] )

= y**2 + 5 x y + 5 x**2

= (137**137 + 1992)(137**137 + 3464)

I presume that you constructed this puzzle by finding
the first two primes > 137**137 which were prime and
equal to either 1 or 4 mod 5.

The smallest product of two distinct primes, = 1 or 4 mod 5,
is 11 * 31 = 341

4**2 = 5 mod 11
6**2 = 5 mod 31

t = 4 mod 11
t = 6 mod 31

37 = 4 mod 11
37 = 6 mod 31

37**2 = 1369
1369 = 5 mod 341

37**2 = 5 mod 341

(5 + sqrt(5))/2 mod 341 = (5 + 37)/2 mod 341 = 21 mod 341

(5 - sqrt(5))/2 MOD 341 = (5 - 37)/2 MOD 341 = 16 MOD 341

Y = 21 X
OR
Y = 16 X

(21 X)**2 + 5 X (21 X) + 5 X**2 = 341

21**2 > 341

(16 X)**2 + 5 x (16x) + 5 x**2 = 341

256 x**2 + 80 x**2 + 5 x**2 = 341

(256 + 80 + 5) x**2 = 341

341 x**2 = 341

x = 1

y = 16

16**2 + 5 * 1 * 16 + 5 * 1**2
= 16**2 + 5 * 16 + 5
= 256 + 80 + 5
= 341.

I expect that 341 is the smallest non-square composite positive integer, M,
such that

there exist x and y
such that

y**2 + 5 x y + 5 x**2 = M.

Thank you Maximilian, for your comment,
which implied that

(137**137 + 1992)*(137**137 + 3464)

is the product of two primes.

>
> there is no other factoring ;)
>
> Maximilian

Kermit
• ... What about 11*19 = 209? quibble,quibble. a.
Message 2 of 2 , Oct 5, 2010
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--- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
>
> > ________________________________________________________________________
> > 1c. Re: qfbsolve exercise
> > Date: Sun Oct 3, 2010 4:24 pm ((PDT))
>
> > Kermit Rose<kermit@> wrote:
> >
> > I expect that 341 is the smallest non-square composite positive integer, M,
> such that
>
> there exist x and y
> such that
>
> y**2 + 5 x y + 5 x**2 = M.
>

What about 11*19 = 209? quibble,quibble.

a.
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