Minimum with natural numbers

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Minimum with natural numbers

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#1 h Apr 10, 2025, 1:51 PM

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Let $x,y,z,t$ be natural numbers such that: $x^2-y^2+t^2=21$ and $x^2+3y^2+4z^2=101$ . Find the min: $M=x^2+y^2+2z^2+t^2$

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#2 h Apr 10, 2025, 2:40 PM

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We have $x^2-y^2\equiv x^2+3y^2+4z^2=101\equiv 1\pmod 4,$ so $x$ is odd and $y$ is even. Say $y=2b$ , with $b\in \mathbb{N}$ , then $x^2+12b^2+4z^2=101,$ so $b=1$ or $b=2$ .
If $b=1$ then $x^2+4z^2=89,$ so $z\leq 4$ , and only $z=4$ gives a solution in natural numbers. In this case we get $(x,y,z)=(5,2,4)$ , which gives $t=0$ , so this doesn't work.
Hence $b=2$ , and so $y=4$ . Then $x^2+4z^2=53,$ so $z\leq 3$ , and only $z=1$ gives a solution in natural numbers. In this case we get $(x,y,z)=(7,4,1)$ , which gives $t^2=-12$ , so this doesn't work.

Hence I assume you mean whole numbers rather than natural numbers.
In this case $b=1$ gives the solution $(x,y,z,t)=(5,2,4,0)$ and $M=61$ .
We also have to check the case $b=0$ . Here we have $x^2+4z^2=101,$ so $z\leq 5$ , and only $z=5$ gives a solution in whole numbers. In this case we get $(x,y,z)=(1,0,5)$ , which gives $t^2=20$ , so this doesn't work.

Hence there are no solutions in natural numbers and only one solution in whole numbers.

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