problems i hate

[zadid xcalibured]

1.prove that any integer can be written as the sum of five integer cubes.
2.prove that every nonnegative integer can be written in the form $a^2+b^2-c^2$ with $a,b,c$ positive integers and $a<b<c$.
how can someone find logic here?

[sourav das]

I'm not angry with this kind of problem, but it really feels bad for me....

I think experience make difference here.....
For 1:

Trying with $(n+1)^3$ , $(n-1)^3$ at first helps

For 2:

try to find (a,b,c) so that $a^2+b^2=c^2$ as then you may work with squares of $ax+1,bx+1,cx+1$ or something like that...

[*Mahi*]

Not logic, but specific forms of squares and cubes.

[zadid xcalibured]

marvellous trick,saurav vai.

[nafistiham]

In these types of problems, the unique trick just does not comes like bolt from the blue.One must pplay with numbers more and more.

[Masum]

1.prove that any integer can be written as the sum of five integer cubes.
2.prove that every nonnegative integer can be written in the form $a^2+b^2-c^2$ with $a,b,c$ positive integers and $a<b<c$.
how can someone find logic here?

Yes, you can.

Not logic, but specific forms of squares and cubes.

Actually I can't agree with you. In the end, when you see the solution to be an explicit one, you may not understand the idea behind this. But in the deep, there is something. Sometimes, a bit investigation and guessing works very well. As I have said in some other posts, I have some favorite parameters to try, and they work very well here. And the crucial idea behind $1$ is, you are not ordered that the cubes must be distinct, and not necessarily positive. So you may choose the numbers as $-a,-a,a+1,a-1$ since
\[(a+1)^3+(a-1)^3=2a^3+6a\]
and this gives \[(a+1)^3+(a-1)^3-a^3-a^3=6a\]
Now we need to work a bit more. Recall that, \[6|n^3-n\] for all $n$. Therefore, we may say that $n^3-n=6a$ for some $a$ and for all $n$. Finally choose $6a=n^3-n$ and you get the final explicit representation. Now write down the formula and note how it looks.
And I explained how to get the version for $2$ on another post.
viewtopic.php?f=26&t=1927

[*Mahi*]

1.prove that any integer can be written as the sum of five integer cubes.
2.prove that every nonnegative integer can be written in the form $a^2+b^2-c^2$ with $a,b,c$ positive integers and $a<b<c$.
how can someone find logic here?

Yes, you can.

Not logic, but specific forms of squares and cubes.

Actually I can't agree with you. In the end, when you see the solution to be an explicit one, you may not understand the idea behind this. But in the deep, there is something. Sometimes, a bit investigation and guessing works very well. As I have said in some other posts, I have some favorite parameters to try, and they work very well here. And the crucial idea behind $1$ is, you are not ordered that the cubes must be distinct, and not necessarily positive. So you may choose the numbers as $-a,-a,a+1,a-1$ since
\[(a+1)^3+(a-1)^3=2a^3+6a\]
and this gives \[(a+1)^3+(a-1)^3-a^3-a^3=6a\]
Now we need to work a bit more. Recall that, \[6|n^3-n\] for all $n$. Therefore, we may say that $n^3-n=6a$ for some $a$ and for all $n$. Finally choose $6a=n^3-n$ and you get the final explicit representation. Now write down the formula and note how it looks.
And I explained how to get the version for $2$ on another post.
viewtopic.php?f=26&t=1927

Thanks for sharing!