Let's see who on /sci/ is actually good at solving problems

>>7665872
What is the question?

Clearly, OP, you are the one here to shitpost. The question isn't even stated in the picture, or your comment. You probably don't even know what the question is.

>>7665872
8=7+1
16=7+9
32=7+25
64=7*9+1
128=7*1+121
256=7*25+81
512=7*49+169
1024=7*9+31^2
...

File: 3face.gif (3MB, 365x272px) Image search: [Google]

3MB, 365x272px

>have proof almost finished
>accidentally click outside of LaTex box
>it's all gone
>mfw
Fuck it, OP can do his own homework.

For small n see >>7665890.
For [math]n \geq 8[/math]:

[eqn]2^{2k} = 7 (5 \cdot 2^{k-4})^2 + (9 \cdot 2^{k-4})^2[/eqn]
[eqn]2^{2k+1} = 7 (7 \cdot 2^{k-4})^2 + (13 \cdot 2^{k-4})^2[/eqn]

>>7665889

To give a proof that the statement in pic related is true.

>>7665890

2,048 17 5
4,096 11 57
8,192 23 67
16,384 45 47
32,768 1 181
65,536 91 87
131,072 89 275
262,144 93 449
524,288 271 101
1,048,576 85 999

>>7665907
Don't x and y need to be odd?

>>7665992
Clearly x and y in that proof are odd. Not sure what you are asking.

>>7666024

>>7666024

In the first equation, [math]x = 5 \cdot 2^{k-4}[/math] is not odd , it's divisible by [math]2^{k-4}[/math]. And [math]y = 9 \cdot 2^{k-4}[/math] is also even. Similarly in the second equation.

There's a factor of a power of 2 so they're even. Or do you think 5*2 is odd just because there's a 5 in it?

Yeah, I know this is probably a troll and you're going to call me autistic or something.

>>7666101
Sorry I was genuinely trying to be helpful and I realized how fucking stupid I am. I need to sleep.

>>7665872
I don't know the answer, but this small amount might help.
(7x-y)(x-y)=7x^2 + y^2 - 8xy, which means
8(2^(n-3) + xy)=2^n + 8xy = (7x-y)(x-y)
because x and y are odd integers, xy is odd, and because 2^(n-3) is even, 2^(n-3) + xy is odd, so 8(2^(n-3) + xy) cannot be divided by 2^4, or powers of 2 greater than 2^4. This means that (7x-y)(x-y) cannot be divided by 2^4 or powers of 2 greater than 2^4, and, because 7x-y and x-y must both be even, exactly one of 7x-y or x-y can be divided by 4, but not by powers of 2 greater than 4, and the other can be divided by 2, but not by powers of 2 greater than 2. So things like x=3 and y=5 can't be solutions because 7x-y = 16, violating this rule.

>>7666300
*2^(n-3) is even for n≥4

Yeah, but 7? Why in the FUCK

Someone show me where to start on this 'proof.'

>>7666300
2^n +8xy should be 2^n - 8xy

>>7666309
It probably involves using induction.

>>7665893
sure :^)

File: 1447719517415.jpg (39KB, 500x500px) Image search: [Google]

39KB, 500x500px

>>7665879
Where are x and y odd?

Learn to read

>>7665872
alright, if you have a solution to 2^n = 7x^2 + y^2, (x is odd, y is odd) then you have a solution to 2^(n+1) because either 2k+x=y exists, where x is odd, y is odd, and k is odd, or 2k-x=y exists, where x is odd, y is odd and k is odd, then 2^n = 7x^2 + (2k+x)^2 = 8x^2 +4(k^2 + kx), which simplifies to 0 = 2x^2 +kx + k^2 - 2^(n-2) (if 2k-x is applicable, then 0 = 2x^2 -kx +k^2 - 2^(n-2)) Applying the quadratic formula (to case 1) we get x = [-k+/-sqrt(k^2 -4(2)(k^2 - 2^(n-2)))]/2(2)=[-k+/-sqrt(2^(n+1)-7k^2)]/4 (in case 2, the quadratic formula gives us x=[k+/-sqrt(2^(n+1)-7k^2)]/4. Because k is odd, we know that sqrt(2^(n+1)-7k^2) is and odd integer, as an even number minus an odd integer is odd, and it is an integer because x is an integer, so, if we have a solution to 2^n=7x^2+y^2, where x and y are both odd integers, then we have a solution to 2^(n+1) = 7x^2 + y^2, where x is odd and y is odd.

File: 1419422137355.png (402KB, 1280x720px) Image search: [Google]

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"This is a very interesting and exceedingly tough problem which was
proposed at the MMO 1985. It is due to Euler, who never published it. It was
taken from his notebook by the proposers. No participant could solve it. It became
a subject of controversy among mathematicians. A prominent number theorist
wrote in the Russian journal Mathematics in School that it was well beyond the
students and required algebraic number theory. I proposed it to our Olympiad
team. One student Eric M¨uller gave a solution after some time, which I did not
understand. I asked him to write it down, so that I could study it in detail. It took
him some time to write it down, since he solved not only this problem but along
with it also over a thousand other problems on 434 pages, all the problems posed
by the trainers in three years. I found the solution of our problem. It was correct."

--Engel: Problem-Solving Strategies (Springer 1998), p.126

>>7666416
so, where's the solution then? since literally no one on /sci/ seems smart enough to solve a thing like this

>>7666428
It's in that book, same page. I can't be arsed to type all that up, sorry. I was simply making the point that trying to come up with a proof while this thread is alive would be pointless. Trying to save some fellow anons some time, you know. If you want to look into the problem, it's a variation of Ramanujan's square equation (google it).

>>766641
What's the k represent?

>>7666411
how do you know k is odd?

>>7666459
2k_1-x=y or 2k_2+x=y
suppose k_1 and k_2 are both even integers (and x and y are both odd integers), then 2k_1-x+2k_2+x=y+y
2(k_1+k_2)=2y
k_1+k_2=y
but if k_1 and k_2 are both even, or both odd, then y is even, but this is a contradiction, so exactly one of k_1 and k_2 is odd (and the other is even).

>>7666428
>>7666411

>>7666416
>It took
>him some time to write it down, since he solved not only this problem but along
>with it also over a thousand other problems on 434 pages, all the problems posed
>by the trainers in three years.
the autism is off the charts

>>7666500
Very nice! Thank you.

>>7666533
no prob

>>7666411
Works great. Nice job anon.

Matlab:

clear all

n=3;
p=2^n;
x=1;
y=1;

disp([n x y 7*x^2+y^2-p]); % disp=display

for n=4:20
if mod(y-x,4)~=0
k=(y-x)/2;
else
k=(y+x)/2;
end
p=p*2;
y=sqrt(p-7*k^2);
x=k;
disp([n x y 7*x^2+y^2-p])
end

Factor 7x^2+y^2 in the Gaussian integers and then use an argument on the factorization of 2^n. It will probably work.

2^3 = 7 + 1
2^4 = 7 + 3^2
If [math]n \ge 3[/math] then, we can write [eqn]2^n = 7(2^{\frac{n-3}{2}})^2 + (2^{\frac{n-3}{2}})^2[/eqn] if n is odd and [eqn]2^n = 2^4(2^{\frac{n-4}{2}})^2 = 7(2^{\frac{n-4}{2}})^2 + (3\cdot 2^{\frac{n-4}{2}})^2[/eqn] if n is even.

>>7666857
isn't x and y even in these cases?

>>7666861
Ah shit you're right, I hadn't noticed the additional requirement

>>7666857
x and y have to be odd

>>7666309
For the case n=3.
Assume is true for n=k
Then solve for n=k+1
Be creative and realize that odd parity is defined as 2n+1 where n belongs to an integer. Since these numbers can be different you will need multiple variables describing x and y.
However, i'm not sure this metherd works as I can't arrive to the solution. I'm stuck on what should be one of the last 3 steps.

>>7665872
I have the proof, however the margins here are too small for me to include it.

Anyone who asks "What is the question?" is a dumb dumb. The question is obviously to prove the statement in pic related.