Indeed, not too difficult: Let $a^2=3x+4y,b^2=4x+3y$. Then $7|a^2+b^2\Rightarrow 7|a,b\Rightarrow 49|a^2,b^2$. On the other hand, $49|4a^2-3b^2=7y\Rightarrow 7|y$ and we're done.

That is exactly what I've done The key step is to know that if $7|a^2+b^2$ then $7|a$ and $7|b$. Pierre.

My solution of problem 3 Let 3x+4y=m ² and 4x+3y=n ² Therefore m ² +n ² =7(x+y) --> m ² +n ² is divided by 7 --> a ² +b ² is divided by 7 where m=7k+a and n=7p+b so a,b=0,1,2…6. We can’t find such values of a and b so that the sum of their squares is divisible by 7 except a=0 and b=0 ( for a=1 trying all the possible values of b: 0,1,2..6 we don’t get the sum of the squares divisible by 7, same thing for a=2,..6). This means that m,n are both divisible by 7. m ² +n ² is divisible by 49 --> x+y is divisible by 7 since m ² +n ² =7(x+y) n ² -m ² =x-y --> x-y is divisible by 49 --> x-y is divisible by 7 So we have x-y, x+y both divisible by 7. therefore the result we are looking for follows.

My solution: Note that the quadratic residues modulo 7 are 0, 1, 2, 4, and thus we immediately see: Lemma. The only residue t modulo 7 such that t and -t are both quadratic residues modulo 7 is t = 0. Now, we have $3x+4y\equiv 3x-3y=3\left( x-y\right)\ \text{mod}\ 7$, and $4x+3y\equiv -3x+3y=-3\left( x-y\right)\ \text{mod}\ 7$. Therefore, both 3 (x - y) mod 7 and -3 (x - y) mod 7 must be quadratic residues modulo 7. Thus, after the Lemma, we must have $3\left( x-y\right) \equiv 0\ \text{mod}\ 7$; hence, $x-y\equiv 0\ \text{mod}\ 7$, and the number x - y is divisible by 7. From $x-y\equiv 0\ \text{mod}\ 7$, it follows that $3x+4y\equiv 3x-3y=3\left( x-y\right) \equiv 0\ \text{mod}\ 7$. Therefore, the number 3x + 4y is divisible by 7. Since this number is a perfect square, it must also be divisible by 49. Similarly, 4x + 3y is divisible by 49. Thus, (3x + 4y) + (4x + 3y) = 7x + 7y = 7 (x + y) is also divisible by 49. This means that x + y is divisible by 7. Since x - y is also divisible by 7, the number 2x = (x + y) + (x - y) must be divisible by 7, too. Since 2 and 7 are coprime, this entails that x is divisible by 7. Similarly, y is divisible by 7. Darij

also here

BTW the following is true: Given any prime $p$ in $4k+3$ form and $a,b$ integers so that $a+b=p$ and $(a,b)=1$. Then any $x,y$ integers which $ax+by,ay+bx$ are both squares then $p|x,y$

7 divides a.a+b.b then 7divides a,b.that means 7 divides x-y.So 7 divides x and y.

let $3x+4y=a^2 . 4x+3y=b^2$. then $7(x+y)=a^2+b^2$. so , 7 divides $a^2+b^2$.checking the Q.R.'s mod7 , we get that 7 divides a , 7 divides b. put a=7m , b=7n. so, $7(x+y)=49(m^2+n^2)$. . so , we get that $x+y=7(m^2+n^2)$. so, 7 divides x+y. also $a^2-b^2=y-x$ , so , 7 divides x-y [ as 7 divides a & b]. so , 7 divides 2x , i.e. 7 divides x & so , 7 divides y.

^^Scientist, $7|x-y$ does not imply $7|x, 7|y$.

pbornsztein wrote: If the positive integers $x$ and $y$ are such that $3x + 4y$ and $4x + 3y$ are both perfect squares, prove that both $x$ and $y$ are both divisible with $7$. We have $3x + 4y = a^2$ and $3y+4x = b^2$ and we get $7(x+y) = a^2+b^2$,$x-y = a^2-b^2$. $a^2$ can be equal $0,1,2,4$ $modulo 7$. Let $k$ $=$ $a^2$ $(mod 7)$ and $l$ $=$ $b^2$ $(mod 7)$ and $k + l$ $=$ $0$ $(mod 7)$. $k = 0$ we get $l = 0$. $k = 1$ no solution. $k = 2$ no solution. $k=4$ no solution. We get the single solution when $a = b = 0$ $(mod 7)$.$x - y = a^2 - b^2$ $\implies x = y$ $(mod 7)$. Now $3x + 4y = 7x = a^2$ $(mod 49)$ and we get $x = y = 0$ $(mod 7)$ because $a^2 = 0$ $(mod 49)$.

Let $3x+4y=a^2, 4x+3y=b^2$. So, $0\equiv 7x+7y\equiv a^2+b^2 \pmod 7$. As $7\equiv 3 \pmod 4$, $7\mid a,b$. $x-y\equiv a^2-b^2\equiv 0 \pmod {49}$. So, $x\equiv y \pmod {49}$. $0\equiv a^2 \equiv 3x+4y\equiv 7x\equiv 7y \pmod {49}$. So, $x\equiv y \equiv 0 \pmod{7}$.