There are two positive integer sets $(b,a),(a,b)$ which satisfy $a^2+b^2=128^2+25^2$ and neither $a$ nor $b$ equals to $128$ and $25$. Find these two sets.
I don't even have a idea how to do this so I would love some help. I have tried to find nearby large numbers such as $130$ and $129$ but both of them don't work. Also will inequality (AM-GM or Cauchy work here?) This is a contest question so I need a preferably fast solution.
Here is one way to proceed:
Render $25^2=5^4, 128^2=4\times 8^4$. Then:
$u^4+4v^4=(u^2-2uv+2v^2)(u^2+2uv+2v^2); u=5, v=8$
$25^2+128^2=5^4+4(8^4)$
$=(5^2-2(5)(8)+2(8^2))(5^2+2(5)(8)+2(8^2))=73\times 233$
Then $73=8^2+3^2$, $233=13^2+8^2$, and we get two product expressions. One gives the original sum of squares and is included as a check, the other gives a second pair:
$(8^2+3^2)(13^2+8^2)$
$=((8\times 13)+(3\times 8))^2+((8\times 8)-(3\times 13))^2=128^2+25^2$
$=((8\times 13)-(3\times 8))^2+((8\times 8)+(3\times 13))^2=80^2+103^2$
Since the factors $73$ and $233$ are prime, this solution is unique apart from trivial operations.
$25^2+128^2$ is of the form
$$a^4+(a^2+b)^2=2a^4+2a^2b+b^2=2(a^2+b)a^2+b^2$$
As luck (or intelligent design) would have it, in this case we have
$$2(a^2+b)=2\cdot128=16^2$$
Thus
$$25^2+128^2=5^4+(5^2+103)=16^2\cdot5^2+103^2=80^2+103^2$$
This doesn't rule out the possibility of a third pair of squares summing to $17009$, but the problem, as stated, doesn't ask for that. If you do want to rule out other possibilities, you can give a proof based on the fact that the only primes less than $\sqrt[3]{17009}\approx25$ congruent to $1$ mod $4$ are $17$, $13$, and $5$, none of which divide $17009$. (It's immediately clear that $5$ and $17$ aren't divisors. As for $13$, I reduced $17009=13000+4009$, then $4009=3900+109$.) Fermat's sum-of-two-squares theory does the rest.
Note, this approach gets an answer without factoring $17009$, but only because we got lucky in that $a^2+b$ in this case is twice a square.
To give credit where credit is due, I got the key idea for this approach from a deleted answer by user Xam.
One may note that by taking both sides modulus $3$, it is clear that $x,y$ must not be a multiple of $3$.
Taking both sides modulus $2$, it is clear either $x$ is odd and $y$ is even or the other way around.
Taking modulus $5$, it is clear that either $x$ is a multiple of $5$ and $y$ is equivalent to $2$ or $3$ modulus $5$.
Putting these hints together, and assuming $x$ is the multiple of $5$, we find
$$x\stackrel?=5,10,20,25,35,40,50,55,65,\dots,130$$
One may also note that if $x<91$, then $y>90$, likewise, if $x>90$, $y<91$.
From here, checking term by term isn't very hard, and one can deduce solutions $(80,103)$.
Since this is a contest question, I think I am doing no harm in showing a solution that is the Gaussian integers version of that of Oscar Lanzi.
$128^2+25^2 = 17009$ factorizes as $73 \cdot 233$. Both primes are congruent to $1$ modulo $4$ so they can be written as the sum of two squares $$ 73 = 8^ 2 + 3^ 2 = (8 + 3 i)(8 - 3 i), \quad 233 = 13^2 + 8^2 = (13 + 8 i) (13 - 8 i). $$ If we group $$ (8 + 3 i)(13 - 8 i) = 128 + 25 i $$ we obtain $17009 = 128^2 + 25^2$. If we group $$ (8 + 3 i)(13 + 8 i) = 80 + 103 i $$ we obtain $17009 = 80^2 + 103^2$.