$a^2 + b^2 =c^4 \text{ }a,b,c\in \Bbb{N}\text{ }a,b,c<100$

Question

The problem is to solve this:
$$a^2 + b^2 =c^4 \text{ }a,b,c\in \Bbb{N}\text{ }a,b,c<100$$ My idea: To see this problem I at first must see idea with Pythagorean triplets, and to problem is to find hypotheses of length square number. Is there any easier approach to this problem?

Answer 1

All Solutions are $$(a=7\land b=24\land c=5)\lor (a=15\land b=20\land c=5)\lor (a=20\land b=15\land c=5)\lor (a=24\land b=7\land c=5)\lor (a=28\land b=96\land c=10)\lor (a=60\land b=80\land c=10)\lor (a=80\land b=60\land c=10)\lor (a=96\land b=28\land c=10)$$

Answer 2

I think, the way with using Pythagorean triplets is the best.

Let $\gcd(a,b,c)=1$.

Thus, there are natural $m$ and $n$ with different parity such that $m>n$ and $\gcd(m,n)=1$ and $a=2mn,$ $b=m^2-n^2$.

Thus, $c^2=m^2+m^2$ and by the same way there are naturals $p$ and $q$ with a different parity, such that $p>q$, $\gcd(p,q)=1$ and $m=2pq$ and $n=p^2-q^2$.

Thus, $a=4pq(p^2-q^2)$, $b=4p^2q^2-(p^2-q^2)^2=6p^2q^2-p^4-q^4$ and $c=p^2+q^2$.

Can you end it now?

For example, for $p=2$ and $q=1$ we obtain: $(a,b,c)=(24,7,5)$.

Answer 3

We need to find Pythagorean triples where the hypotenuse is already a square. We can certainly use $5*(3,4,5)=(15,20,25)$ or $20*(3,4,5)=(60,80,100)$ but we can also use the following formula to find one or more triples for a given hypotenuse, if they exist. For any $C,m$ that yields an integer, we have the $m,n$ we need for a triple with that hypotenuse.

For $C= m^2+n^2,\space n=\sqrt{C-m^2}$ where $\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor$. Let

$$n=\sqrt{25-m^2}\text{ where }\biggl\lceil\sqrt{\frac{25}{2}}\space\space\biggr\rceil=4 \le m\le\bigl\lfloor\sqrt{25}\bigr\rfloor=5$$

We find $n(25,4)=3\rightarrow f(4,3)=(7,24,25)$ but $n(25,5)=0$; there is no triple for $m=5$

We could also find $(3,4,5)$ by using $C=5$, a factor of $25$

$$n=\sqrt{5-m^2}\text{ where }\biggl\lceil\sqrt{\frac{5}{2}}\space\space\biggr\rceil=2 \le m\le\bigl\lfloor\sqrt{5}\bigr\rfloor=2$$ We find $n(5,2)=1\rightarrow f(2,1)=(3,4,5)$

We find no integer values of $n$ for $C=49$ or $C=81$ or their factors

$$n=\sqrt{100-m^2}\text{ where }\biggl\lceil\sqrt{\frac{100}{2}}\space\space\biggr\rceil=8 \le m\le\bigl\lfloor\sqrt{100}\bigr\rfloor=10$$

We find $n(100,8)=6\rightarrow f(8,6)=(28,96,100)$ but $n(100,10)=0$; there is no triple for $m=10$

Answer 4

$a^2 + b^2 =c^4 \quad a,b,c\in \Bbb{N}\quad a,b,c<100$

and

Can you please comment how you get this solutions?

The involved numbers are so small we can compute everything by hand and without brute-force search or such:

Due to $a, b < 100$ we have $c^4=a^2+b^2 < 2\cdot 100^2$ and thus $$c < \sqrt[4]{2}\cdot10<\sqrt{1.44}\cdot10=1.2\cdot10=12$$ where I used $\sqrt 2 < 1.44$. Hence $c$ is 11 at most.

Finding Pythagorean triples is equivalent to solving the norm equation in $\Bbb Z[i]$ with $i=\sqrt{-1}$, the Gaussian integers. As $1\leqslant c\leqslant 11$ we solve the norm equation for primes not 3 mod 4 and multiply the resulting prime ideals together to get the solutions. The only integer primes that decay over $\Bbb Z[i]$ are 2 and 5: $\def\p{\mathfrak p} \def\b{\bar{\p}}$

$2 = \p_2 \b_2$

$5 = \p_5 \b_5$

Real solutions like for for 0 and 1 are of no interest because you only want $a, b\in\Bbb N$. We then take the 4-th powers of these ideals (or their conjugate) to find the solutions for $c$. The only products that do not exceed 11 are: 2, 5 and 2·5=10 where 2 is of no interest because $\p_2^{2n}=2^n$ is real and thus only gives unwanted solutions like $4^2 + 0^2 = 2^4$ (this is because $\p_2=\b_2$ modulo units where $\bar{\,\cdot\,}$ denotes complex conjugation).

We arrive at the following 4 solutions (modulo units and conjugation): \begin{align} s_1 &= \p_5^4 \\ s_2 &= \p_5^3\b_5=5\p_5^2 \\ s_3 &= \p_2^4\p_5^4=2^2\p_5^4 = 4s_1\\ s_4 &= \p_2^4\p_5^3\b_5=2^25\p_5^2 = 4s_2 \end{align} With $\p_5 = 2+i$ we have $\p_5^2=3+4i$ and $\p_5^4=-7+24i$, thus: \begin{array}{rclll} s_1 &=& -7+24i &\Rightarrow& (7,24,5)\\ s_2 &=& 5(3+4i) &\Rightarrow&(15,20,5)\\ s_3 &=& 4(-7+24i)&\Rightarrow&(28,96,10)\\ s_4 &=& 20(3+4i)&\Rightarrow&(60,80,10) \end{array} These solutions are only up to ordering $a$ and $b$, i.e. $(24,7,5)$ etc. are also solutions, of course.