It gets Smaller and Smaller

It is evident that $7|a^{2}+b^{2}$ We see that for any number $a$ , $a^{2} \equiv 0,1,2,4 \pmod{7}$ .So for any two integers a and b , 7 divides $a^{2}+b^{2}$ iff 7 |a & 7|b .So in this case ,let us assume $a = 7a_1$ and $b=7b_1$ .

Hence we have $49(a_1^{2}+b_1^{2}) = 7(c^{2}+d^{2})$ Cancelling 7 from both sides we get $7(a_1^{2}+b_1^{2}) = (c^{2}+d^{2})$ Notice that we get the same equation with $a_1 < a$ and $b_1 < b$ ,we can repeatedly do this as many times as we like to obtain a decreasing sequence of integers but in reality there cannot exist any infinitely decreasing sequence of positive integers as $a,b,c,d$ must be finite...hence we have a contradiction....This method is known as Fermat's Method of Infinite Descent ...

Hence there are $\boxed{0}$ solutions!!

Note that $7\mid a^2+b^2$ . So $7\mid a$ and $7\mid b$ . Assuming there exist some solutions, consider the triplet $a_1,b_1,c_1,d_1$ with smallest value of $a$ . Substituting $a_1=7p$ , $~b_1=7q$ gives:

$a_1^2+b_1^2=49p^2+49q^2=7(c_1^2+d_1^2)~~\Longrightarrow ~ c_1^2+d_1^2=7(p^2+q^2)$

So we have another triplet $(a,b,c,d)=(c_1,d_1,p,q)$ with $c_1 < a$ or $d_1<a$ creating a contradiction. So there exists no solution.

In the note it says "I didn't come up with this problem." So there is no solution for this problem. :P

Let's think that they are Pitágoras Therms. We don't know any triple of Pitágoras Therms with 7 being the hypotenuse. If we had one, or more, we could say that

7(c² + d²) = (a² + b²) -> 7 = x² + y²,

when (x²+y²)(c²+d²)=a²+b². As it doesn't exist, we should say there is no solution.

Any perfect square when divided by 7 gives remainder 0,1,4,2. In the question right hand side is multiple of 7,Hence LHS should also be divisible by 7. if a^{2} gives remainder 1 then b^2 should give 6 which is impossible. Similar is the case with 4 and 2.4 needs 3 and 2 needs 5 both of which are not possible.

It is apparent that the sum of \ $a^\{2\}\\$ and \ $b^\{2\}\\$ must be a multiple of seven. The only way that this can happen is this:

• Case 1: Both squares are the squares of a multiple of seven.

• Case 2: Both squares are not multiples of seven and just happen to add up to a multiple of seven.

Let us examine the second case. First look at the pairs that add to seven. \ $1\+6\\$ , \ $2\+5\\$ , and \ $3\+4\\$ . We see that no matter what we multiply by (remember, it will be distributed to both numbers in the pair), there will never be perfect squares for both pairs. Therefore Case 2 is invalid. For Case 1, both \ $a\\$ and \ $b\\$ must be of the form \ $7x\\$ . Squaring that is \ $a^\{2\}=49x^\{2\}\\$ . Now look at the other side of the equation. By substitution and factoring out 49, \ 49\(x^\{2\}\+y^\{2\} =7 $c^\{2\}\+d^\{2\}\\$ . Dividing by seven, \ 7\(x^\{2\}\+y^\{2\} = $c^\{2\}\+d^\{2\}$ \), which we have already proved to be impossible. Thus, there are \ $\\boxed\{0\}\\$ sets of positive integers that work.