## Make {m^2 + n^2, p^2 + q^2, m*n*p*q} squares

Find four positive integers   $m, \; n, \; p, \; q$   to make the three expressions squares

$m^2 \; + \; n^2$
$p^2 \; + \; q^2$
$m \, n \, p \, q$

math grad - Interest: Number theory
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### 3 Responses to Make {m^2 + n^2, p^2 + q^2, m*n*p*q} squares

1. Paul says:

Here are a few

348^2 + 1015^2 = 353220^2
740^2 + 777^2 = 574980^2
348 X 1015 X 740 X 777 = 450660^2

696^2 + 2030^2 = 1412880^2
1480^2 + 1554^2 = 2299920^2
696 X 2030 X 1480 X 1554 = 1802640^2

1044^2 + 3045^2 = 3178980^2
2220^2 + 2331^2 = 5174820^2
1044 X 3045 X 2220 X 2331 = 4055940^2

435^2 + 3248^2 = 1412880^2
2260^2 + 2373^2 = 5362980^2
435 X 3248 X 2260 X 2373 = 2752680^2

555^2 + 4144^2 = 2299920^2
1356^2 + 3955^2 = 5362980^2
555 X 4144 X 1356 X 3955 = 3512040^2

1392^2 + 4060^2 = 5651520^2
2960^2 + 3108^2 = 9199680^2
1392 X 4060 X 2960 X 3108 = 7210560^2

803^2 + 4380^2 = 3517140^2
2928^2 + 3355^2 = 9823440^2
803 X 4380 X 2928 X 3355 = 5877960^2

1740^2 + 5075^2 = 8830500^2
3700^2 + 3885^2 = 14374500^2
1740 X 5075 X 3700 X 3885 = 11266500^2

245^2 + 6000^2 = 1470000^2
3603^2 + 4804^2 = 17308812^2
245 X 6000 X 3603 X 4804 = 5044200^2

459^2 + 6188^2 = 2840292^2
949^2 + 6132^2 = 5819268^2
459 X 6188 X 949 X 6132 = 4065516^2

2088^2 + 6090^2 = 12715920^2
4440^2 + 4662^2 = 20699280^2
2088 X 6090 X 4440 X 4662 = 16223760^2

870^2 + 6496^2 = 5651520^2
4520^2 + 4746^2 = 21451920^2
870 X 6496 X 4520 X 4746 = 11010720^2

2436^2 + 7105^2 = 17307780^2
5180^2 + 5439^2 = 28174020^2
2436 X 7105 X 5180 X 5439 = 22082340^2

1110^2 + 8288^2 = 9199680^2
2712^2 + 7910^2 = 21451920^2
1110 X 8288 X 2712 X 7910 = 14048160^2

2784^2 + 8120^2 = 22606080^2
5920^2 + 6216^2 = 36798720^2
2784 X 8120 X 5920 X 6216 = 28842240^2

1606^2 + 8760^2 = 14068560^2
5856^2 + 6710^2 = 39293760^2
1606 X 8760 X 5856 X 6710 = 23511840^2

3132^2 + 9135^2 = 28610820^2
6660^2 + 6993^2 = 46573380^2
3132 X 9135 X 6660 X 6993 = 36503460^2

1305^2 + 9744^2 = 12715920^2
6780^2 + 7119^2 = 48266820^2
1305 X 9744 X 6780 X 7119 = 24774120^2

Paul.

• benvitalis says:

2. Paul says:

No parametric solution, I solved for Pythagorean triples for equal hypotenuse, then took all 2 length subsets and tested for m n p q equal to a square.

The first solution is when the hypotenuse is 1073. here my MMa code

```Clear[r,s,n];
Monitor[Do[p=Solve[r^2+s^2==n^2&&r>0&&s>0&&r<s,{r,s},Integers];
p={r,s}/.p;t=Cases[Subsets[p,{2}],{{a_,b_},{c_,d_}}/;
IntegerQ[Sqrt[a b c d]]];t=Flatten[t,1];If[Length[t]>0,
Print[t[[1,1]],"^2 + ",t[[1,2]],"^2 = ",Sqrt[t[[1,1]]^2 t[[1,2]]^2],"^2"];
Print[t[[2,1]],"^2 + ",t[[2,2]],"^2 = ",Sqrt[t[[2,1]]^2 t[[2,2]]^2],"^2"];
Print[t[[1,1]]," X ",t[[1,2]]," X ",t[[2,1]]," X ",t[[2,2]]," = ",Sqrt[t[[1,1]] t[[1,2]] t[[2,1]] t[[2,2]]],"^2"];Print[]],{n,1073,10000}],n]```

Paul.