Let $a, b, c$ and $d$ be prime numbers such that $a>3 b>6 c>12 d$ and $a^{2}-b^{2}+c^{2}-d^{2}=1749$. Determine all possible values of $a^{2}+b^{2}+c^{2}+d^{2}$.

## Solutions
Answer: the only possible value is 1999 .

Since $a^{2}-b^{2}+c^{2}-d^{2}$ is odd, one of the primes $a, b, c$ and $d$ must be 2 , and in view of $a>3 b>6 c>12 d$ we must have $d=2$. Now

$$
1749=a^{2}-b^{2}+c^{2}-d^{2}>9 b^{2}-b^{2}+4 d^{2}-d^{2}=8 b^{2}-12,
$$

implying $b \leqslant 13$. From $4<c<\frac{b}{2}$ we now have $c=5$ and $b$ must be either 11 or 13 . It remains to check that $1749+2^{2}-5^{2}+13^{2}=1897$ is not a square of an integer, and $1749+2^{2}-5^{2}+11^{2}=1849=43^{2}$. Hence $b=11$, $a=43$ and

$$
a^{2}+b^{2}+c^{2}+d^{2}=43^{2}+11^{2}+5^{2}+2^{2}=1999
$$