可能是一个素数公式

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(1105846759997316104484129523097^2+3)/(2^101-1)＝482347839529437263004938742412

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(1176289929542023307880427126232^2+3)/(2^103-1)＝136439212636043267685016099261

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(70834587215330078556119102205056^2+3)/(2^107-1)＝30922969977534372035774906613757

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(78676610129673952743199618487727214612^2+3)/(2^127-1)＝36381602946443044734029544940562775661

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(1352991471484932553120430861625628626501148470238064320141651605715112580856244046791115838201919442236187723639505287050784484963256584245667191556872779379^2+3)/(2^521-1)＝266662764518558921419032713972300564529835303454288748091778629747202402045135390830716896661815639643323095881579431584756113716893612545416911231473961044

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(32945091792165156302706874241662114570936374279843225723757112572746476916601440173115592619992767634279364553001054861891738523234338439858321192427101294194679756931491642762145212^2+3)/(2^607-1)＝2043497335670024072264687077088303423909408317200097383373377945849132131695145615147087302846974669554016679142974932279024818541198946763034399224883122344277106421071610652679661

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(54257397584226315303852092^2+3)/(2^89-1)＝4756070729099607400034797

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\(\frac{a^2+3}{c}＝2^k-1\)，素数\(k＞0\)，数学软件求出\(a\)和\(c\)值

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已知:整数\(a＞0\)，\(c＞0\)，\(\frac{a^2+3}{c}＝2^k-1\)，素数\(k＞0\)，\(p＞0\)
方程\(\frac{a^2+3}{c}-2^k+1＝0\)，最多两个整数解
结论:\(2^k-1＝p\)

太阳

数学软件强大，找到梅森素数1亿位难度不大