As already stated in title, find 7-tuples ($a_1,a_2,a_3,a_4,b_1,b_2,b_3$) of pairwise distinct positive integers such that
$$a_1^2+a_2^2+a_3^2+a_4^2=b_1^2+b_2^2+b_3^2$$
This came in RMO 2016 Delhi paper where one was asked to prove that infinite such tuples exist. I have no idea how to do so.
On
Rewrite as $$ \begin{align}a_4^2&=b_1^2-a_1^2+b_2^2-a_2^2+b_3^2-a_3^2\\ &=(b_1-a_1)(b_1+a_1)+(b_1-a_2)(b_1+a_2)+(b_3-a_3)(b_3+a_3)\end{align}$$ Thus we are essentially asked to write a square as sum of three composite numbers (with mild additional conditions), for example $$100=3\cdot 11+5\cdot 7+2\cdot 16$$ gives us $$4^2+1^2+\color{red}{7^2}+10^2=\color{red}{7^2}+6^2+9^2 $$ as an almost solution. The main problem with finding a solution is that if you start with a too small square on the left, you may run into a prime after subtracting two suitable composites (primes are very common among small numbers) are some of the numbers coincide (as above); this is easily avoided by starting with a larger left hand side. For example, we can try $$121=3\cdot 11+4\cdot 6+2\cdot 32,$$ which gives us $$ 4^2+1^2+15^2+11^2=7^2+5^2+17^2$$ and from this we get one set of infinitely many solutions $$(4n,n,15n,11n,7n,5n,17n),\quad n\in\Bbb N $$
On
For Diophantine equation.
$$a^2+b^2+c^2+d^2=x^2+y^2+z^2$$
You can record a parameterization.
$$a=2(p+s+r-t-q)k$$
$$b=k^2+t^2+q^2-p^2-s^2-r^2$$
$$c=p^2+s^2+r^2+t^2-k^2-q^2-2(p+s+r-q)t$$
$$d=p^2+s^2+r^2+q^2-k^2-t^2-2(p+s+r-t)q$$
$$x=p^2+k^2+t^2+q^2-s^2-r^2+2(s+r-t-q)p$$
$$y=s^2+k^2+t^2+q^2-p^2-r^2+2(p+r-t-q)s$$
$$z=r^2+k^2+t^2+q^2-p^2-s^2+2(p+s-t-q)r$$
One approach is to find a solution by trial. We can start by finding two pairs of squares that sum to the same number, then adding in a Pythagorean triple. So $39^2+52^2=65^2=25^2+60^2$ and a solution is $3^2+4^2+39^2+52^2=5^2+25^2+60^2$. Now replace the $3^2+4^4=5^2$ with other triangles. As there are an infinite number of primitive Pythagorean triangles, this will give an infinite set of solutions that have no common factor.